MATH 114 - QUANTITATIVE
REASONING - Systems of Equations
Question Bank - Set 2
Liberty University
Question 1
Question
Solve the following system of equations:
(2x+ 3y= 5
x2+y2= 10
Solution
Step 1: Multiply the first equation by 2 to simplify the system:
(4x+ 6y= 10
x2+y2= 10
Step 2: Subtract the second equation from the first equation to eliminate y:
4x+ 6y−(x2+y2) = 10 −10
4x+ 6y−x2−y2= 0
4x−x2+ 6y−y2= 0
x2−4x+y2−6y= 0
(x−2)2+ (y−3)2= 13
Step 3: Now we have the system
((x−2)2+ (y−3)2= 13
x2+y2= 10
We can solve this system by substituting (x, y) = (2 + √2 cos(t),3 + √13 sin(t)),
where tis a parameter.
Thus, the solutions to the system are
(x, y) = 2 + √2 cos(t),3 + √13 sin(t)
Question 2
Question
Solve the following system of equations:
(x2+y2= 25
x+y= 7
Solution
Step 1: We recognize the second equation as the equation of a line in slope-
intercept form. We can rewrite it as y=−x+ 7.
Step 2: Next, we substitute y=−x+ 7 into the first equation x2+y2= 25:
x2+ (−x+ 7)2= 25
Step 3: Expand and simplify the expression on the left side:
x2+ (x2−14x+ 49) = 25
Step 4: Combine like terms:
2x2−14x+ 49 = 25
Step 5: Rearrange the equation into standard form:
2x2−14x+ 24 = 0
Step 6: Divide the entire equation by 2 to simplify:
x2−7x+ 12 = 0
Step 7: Factor the quadratic equation:
(x−3)(x−4) = 0
Step 8: Set each factor equal to zero and solve for x:
x−3 = 0 =⇒x= 3
x−4 = 0 =⇒x= 4
2
Step 9: Substitute x= 3 and x= 4 back into the linear equation y=−x+ 7
to find the corresponding yvalues: For x= 3:
y=−(3) + 7 = 4
For x= 4:
y=−(4) + 7 = 3
Step 10: Therefore the solution to the system of equations is:
(x, y) = {(3,4),(4,3)}
Question 3
Question
Find the solution to the following system of equations:
(x2+y2= 25
xy = 12
Solution
Step 1: We can start by squaring the second equation and simplifying to elimi-
nate y2:
(x2+y2)2= (25)2
x4+ 2x2y2+y4= 625
x4+ 2(12)2+y4= 625
x4+ 288 + y4= 625
x4+y4= 337
Step 2: Now, we substitute x2+y2= 25 and x4+y4= 337 into the equations:
(x2+y2= 25
x4+y4= 337
Step 3: Let’s denote x2as aand y2as b. The system now becomes:
(a+b= 25
a2+b2= 337
Step 4: We can solve this new system of equations to find the values of a
and b: From a+b= 25, we can express ain terms of bas a= 25 −b.
Step 5: Substitute a= 25 −binto a2+b2= 337:
(25 −b)2+b2= 337
3
625 −50b+b2+b2= 337
2b2−50b+ 288 = 0
b2−25b+ 144 = 0
(b−9)(b−16) = 0
Step 6: We find that b= 9 or b= 16. Thus, the corresponding values
for aare 16 and 9 respectively. Therefore, the possible solutions for (x, y) are
(−3,−4),(−4,−3),(3,4),and (4,3).
Question 4
Question
Find the solution to the system of equations:
(2x−3y= 5
3x+ 4y= 11
Solution
Step 1: Multiply the first equation by 3 and the second equation by 2 to make
the coefficients of xin both equations equal.
(6x−9y= 15
6x+ 8y= 22
Step 2: Subtract the first equation from the second equation to eliminate x.
17y= 7
Step 3: Solve for y.
y=7
17
Step 4: Substitute y=7
17 back into the first equation and solve for x.
2x−37
17= 5
2x−21
17 = 5
2x=104
17
x=52
17
Therefore, the solution to the system of equations is x=52
17 and y=7
17 .
4
Question 5
Question
Solve the following system of equations:
2x+ 4y−z= 5
x−3y+ 2z=−8
3x+ 2y−5z= 10
Solution
Step 1: Multiply the second equation by 2 and add it to the first equation to
eliminate x:
2x+ 4y−z= 5
4x−6y+ 4z=−16
3x+ 2y−5z= 10
Step 2: Multiply the first equation by 2 and subtract it from the third
equation to eliminate x:
2x+ 4y−z= 5
4x−6y+ 4z=−16
−5y+ 3z= 0
Step 3: Solve the third equation for yin terms of z:
−5y+ 3z= 0 =⇒y=3z
5
Step 4: Substitute y=3z
5back into the first equation:
2x+ 4 3z
5−z= 5 =⇒2x+12z
5−z= 5 =⇒2x+7z
5= 5
Step 5: Multiply the above equation by 5 to eliminate the fraction:
10x+ 7z= 25
Step 6: Substitute y=3z
5back into the second equation:
x−33z
5+ 2z=−8 =⇒x−9z
5+ 2z=−8 =⇒x−4z
5=−8
Step 7: Multiply the above equation by 5 to eliminate the fraction:
5x−4z=−40
5
Step 8: Now, we have a system of two equations:
(10x+ 7z= 25
5x−4z=−40
Step 9: Solve the above system of equations to find xand z.
(10x+ 7z)−2(5x−4z) = 25 −2(−40)
10x+ 7z−10x+ 8z= 25 + 80
15z= 105
z= 7
Step 10: Substitute z= 7 back into the equation 5x−4z=−40 to find x:
5x−4(7) = −40 =⇒5x−28 = −40 =⇒5x=−12 =⇒x=−2.4
Step 11: Substitute x=−2.4 and z= 7 back into the equation 2x+4y−z= 5
to find y:
2(−2.4) + 4y−7 = 5 =⇒ −4.8+4y−7 = 5 =⇒4y= 16.8 =⇒y= 4.2
Thus, the solution to the system of equations is:
x=−2.4, y = 4.2, z = 7
Question 6
Question
Solve the following system of equations:
(2x+ 3y= 8
4x−3y= 2
Solution
Step 1: Let’s write the system of equations:
(2x+ 3y= 8
4x−3y= 2
Step 2: Add the two equations together to eliminate y:
(2x+ 3y) + (4x−3y) = 8 + 2
6
Step 3: Simplify the equation:
6x= 10
Step 4: Solve for x:
x=10
6=5
3
Step 5: Substitute xback into one of the original equations. Let’s use the
first equation:
25
3+ 3y= 8
Step 6: Simplify the equation and solve for y:
10
3+ 3y= 8
3y= 8 −10
3
3y=24 −10
3
3y=14
3
y=14
9
Step 7: The solution to the system of equations is x=5
3and y=14
9.
Question 7
Question
Solve the following system of equations:
3x−2y+z= 7
x+ 4y−2z=−4
2x−3y+ 2z= 5
Solution
Step 1: Begin by expressing the system of equations as an augmented matrix.
3−2 1 7
1 4 −2−4
2−3 2 5
7
Step 2: Use row operations to transform the matrix to row-echelon form.
3−2 1 7
1 4 −2−4
2−3 2 5
→
1 4 −2−4
010
3−5
3
25
3
0−11 5 13
Step 3: Continue row operations to reduce the matrix to row-echelon form.
1 4 −2−4
010
3−5
3
25
3
0−11 5 13
→
1 4 −2−4
0 1 −1
2
5
2
0−11 5 13
→
1 4 −2−4
0 1 −1
2
5
2
0 0 0 0
Step 4: Interpret the row-echelon form of the system. The third equation
0 = 0 indicates that there are infinitely many solutions. The system has one
free variable, which can be denoted as t. Now express x,y, and zin terms of t.
x=−4−4t,y=5
2+1
2t,z=t
So, the solution to the system of equations is:
x=−4−4t
y=5
2+1
2t
z=t
Question 8
Question
Solve the following system of equations:
(x2+y2= 25
xy = 12
Solution
Step 1: From the second equation, we can express yas 12
x. Substituting this
into the first equation, we get:
x2+12
x2
= 25
Step 2: Simplifying the equation gives:
x2+144
x2= 25
Step 3: Multiplying through by x2gives us a quadratic equation:
x4−25x2+ 144 = 0
8
Step 4: This equation can be factored as (x2−9)(x2−16) = 0, which gives
us two sets of solutions: x=±3 and x=±4.
Step 5: We can now find the corresponding values of yby using the second
equation. For x= 3: y=12
3= 4 or y=−4 For x=−3: y=12
−3=−4 or y= 4
For x= 4: y=12
4= 3 or y=−3 For x=−4: y=12
−4=−3 or y= 3
Therefore, the solutions to the system are:
(x, y) = (3,4),(−3,−4),(4,3),(−4,−3)
Question 9
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 5
Solution
Step 1: We can start solving this system of equations by eliminating one of the
variables. Let’s eliminate y: Step 2: Multiply the second equation by 3 to make
the coefficients of yequal:
(2x+ 3y= 7
12x−3y= 15
Step 3: Add the two equations together to eliminate y:
14x= 22
Step 4: Solve for xby dividing both sides by 14:
x=22
14 =11
7
Step 5: Substitute x=11
7back into the first equation to solve for y:
211
7+ 3y= 7
22
7+ 3y= 7
Step 6: Subtract 22
7from both sides:
3y= 7 −22
7=35
7−22
7=13
7
9
Step 7: Solve for yby dividing both sides by 3:
y=13
21
Step 8: The solution to the system of equations is x=11
7and y=13
21 .
Question 10
Question
Solve the following system of equations:
(3x−2y= 5
2x2+ 5y=−1
Solution
Step 1: We can start by solving the first equation for y.
3x−2y= 5 =⇒ −2y=−3x+ 5 =⇒y=3
2x−5
2
Step 2: Substitute yin terms of xinto the second equation and solve for x.
2x2+ 5 3
2x−5
2=−1
Simplify the equation:
2x2+15
2x−25
2=−1
4x2+ 15x−25 = −2
4x2+ 15x−23 = 0
Step 3: Use the quadratic formula to find the solutions for x. The solutions
are:
x=−15 ±p152−4∗4∗(−23)
2∗4
x=−15 ±√225 + 368
8
x=−15 ±√593
8
Thus, x=−15+√593
8or x=−15−√593
8.
Step 4: Substitute each value of xback into y=3
2x−5
2to find the respective
values of y. Therefore, the solution to the system of equations is:
−15 + √593
8,7−3√593
8!and −15 −√593
8,7+3√593
8!
10
Question 11
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 5
Solution
Step 1: Solve the second equation for yin terms of x.
4x−y= 5
y= 4x−5
Step 2: Substitute yfrom the second equation into the first equation.
2x+ 3(4x−5) = 7
2x+ 12x−15 = 7
14x−15 = 7
Step 3: Solve for x.
14x= 22
x=22
14
x=11
7
Step 4: Substitute the value of xback into the equation y= 4x−5 to find
y.
y= 4 11
7−5
y=44
7−5
y=44
7−35
7
y=9
7
Step 5: The solution to the system of equations is x=11
7and y=9
7.
Question 12
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 5
11
Solution
Step 1: Solve the second equation for y.
4x−y= 5
y= 4x−5
Step 2: Substitute y= 4x−5 into the first equation and solve for x.
2x+ 3(4x−5) = 7
2x+ 12x−15 = 7
14x−15 = 7
14x= 22
x=22
14
x=11
7
Step 3: Substitute x=11
7into y= 4x−5 and solve for y.
y= 4 11
7−5
y=44
7−5
y=44
7−35
7
y=9
7
Therefore, the solution to the system of equations is x=11
7and y=9
7.
Question 13
Question
Solve the following system of equations:
(2x+ 3y= 7
3x−2y= 8
Solution
Step 1: Solve the first equation for x.
2x+ 3y= 7 =⇒x=7−3y
2
12
Step 2: Substitute the expression for xinto the second equation.
37−3y
2−2y= 8
Step 3: Simplify the equation.
21 −9y
2−2y= 8
Step 4: Multiply through by 2 to clear the fractions.
21 −9y−4y= 16
Step 5: Combine like terms.
21 −13y= 16
Step 6: Solve for y.
−13y=−5 =⇒y=5
13
Step 7: Substitute the value of yback into the equation to find x.
x=7−35
13
2
Step 8: Simplify to find the value of x.
x=7−15
13
2=7·13 −15
2·13 =91 −15
26 =76
26 =38
13
Therefore, the solution to the system of equations is x=38
13 and y=5
13 .
Question 14
Question
Solve the following system of equations:
(x2−2xy +y2= 25
x2+ 2xy +y2= 49
Solution
Step 1: Start by subtracting the second equation from the first equation to
eliminate 2xy.
x2−2xy +y2= 25
x2+ 2xy +y2= 49
−4xy =−24
xy = 6
13
Step 2: Next, we can rewrite the original first equation in terms of xy.
(x+y)2=x2+ 2xy +y2= 49
x+y=√49
x+y= 7 (since x, y are positive)
Step 3: Now we have a system of equations:
(x+y= 7
xy = 6
We can solve this system by substituting y= 7 −xinto the second equation.
x(7 −x) = 6
7x−x2= 6
x2−7x+ 6 = 0
(x−1)(x−6) = 0
Step 4: From here, we have two possible solutions:
x= 1, y = 6
x= 6, y = 1
Therefore, the solutions to the system of equations are x= 1, y = 6 and
x= 6, y = 1.
Question 15
Question
Solve the following system of equations:
(2x+ 3y= 11
3x−2y= 1
Solution
Step 1: We will solve the system of equations using the method of substitution.
We can rearrange the first equation to solve for xin terms of y:
2x+ 3y= 11 ⇒2x= 11 −3y⇒x=11 −3y
2
Step 2: Substitute the expression for xinto the second equation:
311 −3y
2−2y= 1
14
Step 3: Simplify the equation and solve for y:
33 −9y
2−2y= 1 ⇒33−9y−4y= 2 ⇒33−13y= 2 ⇒ −13y=−31 ⇒y=31
13 =31
13
Step 4: Substitute yback into the expression for xto solve for x:
x=11 −3·31
13
2=11 −93/13
2=11 ·13 −93
13 ·2=143 −93
26 =50
26 =25
13
Step 5: The solution to the system of equations is x=25
13 and y=31
13 .
Question 16
Question
Solve the following system of equations for xand y:
(3x+ 2y= 7
5x−4y= 1
Solution
Step 1: Multiply the first equation by 2 and the second equation by 3 to elimi-
nate y:
(6x+ 4y= 14
15x−12y= 3
Step 2: Add the two equations together to eliminate y:
21x= 17
Step 3: Solve for x:
x=17
21
Step 4: Substitute xback into the first equation to solve for y:
317
21+ 2y= 7
17 + 2y= 7
2y=−10
y=−5
Therefore, the solution to the system of equations is x=17
21 and y=−5.
15
Question 17
Question
Solve the following system of equations:
(3x−2y= 4
x2+y2= 25
Solution
Step 1: Solve the first equation for xin terms of y.
3x−2y= 4
3x= 2y+ 4
x=2y+ 4
3
Step 2: Substitute xinto the second equation.
2y+ 4
32
+y2= 25
4y2+ 16y+ 16
9+y2= 25
4y2+ 16y+ 16 + 9y2= 225
13y2+ 16y−209 = 0
Step 3: Solve the quadratic equation for y.
y=−b±√b2−4ac
2a
y=−16 ±p162−4(13)(−209)
2(13)
y=−16 ±√256 + 10868
26
y=−16 ±√11124
26
y=−16 ±106
26 (since √11124 = 106)
y=90
26 or y=−122
26
y=45
13 or y=−61
13
16
Step 4: Substitute the yvalues back to solve for x.
x=2(45
13 )+4
3=94
13
x=2(−61
13 )+4
3=−34
13
Therefore, the solutions to the system of equations are x=94
13 ,y=45
13 and
x=−34
13 ,y=−61
13 .
Question 18
Question
Solve the following system of equations:
(3x−2y= 4
2x+y= 1
Solution
Step 1: We can solve this system of equations using either substitution or elim-
ination method. Let’s use the elimination method.
Step 2: Multiply the second equation by 2:
(3x−2y= 4
4x+ 2y= 2
Step 3: Add the two equations together to eliminate y:
7x= 6
Step 4: Solve for x:
x=6
7
Step 5: Substitute the value of xback into the second equation and solve
for y:
26
7+y= 1
y= 1 −12
7
y=−5
7
Step 6: Therefore, the solution to the given system of equations is:
x=6
7, y =−5
7
17
Question 19
Question
Solve the following system of equations:
(2x+ 3y= 10
4x−y= 5
Solution
Step 1: Solve the second equation for yin terms of x.
4x−y= 5
y= 4x−5
Step 2: Substitute y= 4x−5 into the first equation and solve for x.
2x+ 3(4x−5) = 10
2x+ 12x−15 = 10
14x−15 = 10
14x= 25
x=25
14
Step 3: Substitute x=25
14 back into y= 4x−5 and solve for y.
y= 4 25
14−5
y=100
14 −70
14
y=30
14
y=15
7
Therefore, the solution to the system of equations is x=25
14 and y=15
7.
Question 20
Question
Find the solution to the following system of equations:
(3x+ 4y= 5
2x−y= 8
18
Solution
Step 1: Let’s rewrite the system of equations in standard form:
3x+ 4y= 5
2x−y= 8
Step 2: We will use the elimination method to solve the system. To eliminate
y, we multiply equation (2) by 4 and add it to equation (1):
3x+ 4y= 5
8x−4y= 32
Step 3: Adding the two equations we get:
11x= 37
Step 4: Solve for x:
x=37
11
Step 5: Substitute xback into equation (2) to solve for y:
237
11−y= 8
Step 6: Simplify and solve for y:
y=1
11
Step 7: Therefore, the solution to the system of equations is x=37
11 and
y=1
11 .
Question 21
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 5
Solution
Step 1: Solve the second equation for y.
4x−y= 5
⇒y= 4x−5
19
Step 2: Substitute y= 4x−5 into the first equation.
2x+ 3(4x−5) = 7
2x+ 12x−15 = 7
14x−15 = 7
14x= 22
x=22
14
x=11
7
Step 3: Substitute x=11
7into y= 4x−5 to solve for y.
y= 4 11
7−5
y=44
7−5
y=44
7−35
7
y=9
7
So, the solution to the system of equations is x=11
7and y=9
7.
Question 22
Question
Consider the following system of equations:
(3x−y= 5
x2−2xy +y2= 9
Solve the system of equations.
Solution
Step 1: We will start by solving the first equation for yin terms of x.
3x−y= 5
y= 3x−5
20
Step 2: Substituting y= 3x−5 into the second equation, we get:
x2−2x(3x−5) + (3x−5)2= 9
x2−6x2+ 10x+ 9x2−30x+ 25 = 9
−2x2−20x+ 25 = 9
−2x2−20x+ 16 = 0
x2+ 10x−8 = 0
Step 3: Next, we’ll solve the quadratic equation x2+ 10x−8 = 0 using the
quadratic formula:
x=−b±√b2−4ac
2a
x=−10 ±p102−4(1)(−8)
2(1)
x=−10 ±√100 + 32
2
x=−10 ±√132
2
x=−10 ±2√33
2
x=−5±√33
So, the solutions for xare x1=−5 + √33 and x2=−5−√33.
Step 4: Using the values of x1and x2to find the corresponding yvalues:
y1= 3(−5 + √33) −5 = −15 + 3√33 −5 = −20 + 3√33
y2= 3(−5−√33) −5 = −15 −3√33 −5 = −20 −3√33
Therefore, the solutions to the system of equations are (−5 + √33,−20 +
3√33) and (−5−√33,−20 −3√33).
Question 23
Question
Find the solution to the system of equations:
(2x+ 3y= 7
4x−3y= 5
21
Solution
Step 1: Add the two equations together to eliminate y:
(2x+ 3y) + (4x−3y) = 7 + 5
6x= 12
x= 2
Step 2: Substitute x= 2 back into the first equation to solve for y:
2(2) + 3y= 7
4+3y= 7
3y= 3
y= 1
Step 3: The solution to the system of equations is x= 2 and y= 1.
Question 24
Question
Solve the following system of equations:
3x−y+z= 2
2x+y−2z= 3
x+ 3y−4z=−1
Solution
Step 1: We will use the method of substitution to solve this system of equations.
Let’s start by solving the first equation for y: From the first equation:
y= 3x+z−2
Step 2: Next, we substitute y= 3x+z−2 into the second and third equations
to eliminate y: Substitute y= 3x+z−2 into the second equation:
2x+ (3x+z−2) −2z= 3
Simplify to get:
5x+z−2z= 5
5x−z= 5 (Equation 4)
Step 3: Substitute y= 3x+z−2 into the third equation:
x+ 3(3x+z−2) −4z=−1
22
Simplify to get:
x+ 9x+ 3z−6−4z=−1
10x−z= 5 (Equation 5)
Step 4: Now, we will solve Equations 4 and 5 simultaneously. Multiply
Equation 4 by 2:
2(5x−z) = 2(5)
10x−2z= 10 (Equation 6)
Step 5: Subtract Equation 5 from Equation 6:
(10x−2z)−(10x−z) = 10 −5
−x= 5
x=−5
Step 6: Substitute x=−5 into Equation 4:
5(−5) −z= 5
−25 −z= 5
z=−30
Step 7: Substitute x=−5 and z=−30 into Equation 1 to solve for y:
3(−5) −y+ (−30) = 2
−15 −y−30 = 2
y=−13
Therefore, the solution to the system of equations is:
x=−5
y=−13
z=−30
Question 25
Question
Solve the system of equations:
(4x−3y= 5
2x2+ 3xy = 1
23
Solution
Step 1: Let’s start by multiplying the first equation by 2 to eliminate ywhen
adding the equations. Step 2: We will then substitute the value of yback into
the first equation to solve for x. Step 3: Finally, we will substitute the value of
xback into the first equation to solve for y.
Step 1: Multiply the first equation by 2:
2(4x−3y) = 2(5)
8x−6y= 10
Step 2: Substitute 8x−6yinto the second equation:
2x2+ 3x(8x−6y)=1
2x2+ 24x2−18xy = 1
26x2−18xy = 1
Step 3: Substitute yback into the first equation:
4x−326x2−1
18x= 5
4x−78x2−3
18 = 5
72x2−54x−78x2+ 3 = 90
−6x2−54x+ 3 = 90
−6x2−54x−87 = 0
2x2+ 18x+ 29 = 0
This is a quadratic equation, so we can apply the quadratic formula to solve
for x:
x=−18 ±p182−4(2)(29)
2(2)
x=−18 ±√324 −232
4
x=−18 ±√92
4
x=−18 ±2√23
4
x=−9
2±√23
2
Therefore, the solutions for the system of equations are:
x=−9
2+√23
2
y=−5
3−5√23
18
24
Question 26
Question
Solve the following system of equations:
(3x+ 2y= 5
6x+ 3y= 8
Solution
Step 1: Multiply the first equation by 2 to make the coefficients of ythe same:
(6x+ 4y= 10
6x+ 3y= 8
Step 2: Subtract the second equation from the first equation:
(6x+ 4y)−(6x+ 3y) = 10 −8
y= 2
Step 3: Substitute y= 2 back into the first equation to solve for x:
3x+ 2(2) = 5
3x+ 4 = 5
3x= 1
x=1
3
So the solution to the system of equations is x=1
3and y= 2.
Question 27
Question
Solve the following system of equations:
(2x+ 3y= 7
x2+y2= 10
Solution
To solve this system of equations, we will first solve for one variable in terms of
the other and then substitute back to find the values of both variables.
25
Step 1: Solve for xin terms of yfrom the first equation:
2x+ 3y= 7
2x= 7 −3y
x=7
2−3
2y
Step 2: Substitute xinto the second equation:
7
2−3
2y2
+y2= 10
49
4−21
2y+9
4y2+y2= 10
49
4−21
2y+13
4y2= 10
52 −42y+ 13y2= 40
13y2−42y+ 12 = 0
Step 3: Solve the quadratic equation: Solving the quadratic equation
using the quadratic formula:
y=42 ±p422−4(13)(12)
2(13)
y=42 ±√1764 −624
26
y=42 ±√1140
26
y=42 ±2√285
26
y=21 ±√285
13
Therefore, the solutions to the system of equations are:
(x=7
2−3
221±√285
13
y=21±√285
13
Question 28
Question
Find all solutions to the following system of equations:
2x+ 3y−z= 4
x−2y+ 3z=−2
3x+ 2y+ 4z= 10
26
Solution
Step 1: We will solve the system using the method of substitution.
Step 2: From the first equation, we can isolate z:
z= 2x+ 3y−4
Step 3: Substituting zin the second equation, we get:
x−2y+ 3(2x+ 3y−4) = −2
Step 4: Simplifying the equation, we obtain:
x−2y+ 6x+ 9y−12 = −2
7x+ 7y= 10
x+y=10
7
Step 5: Substituting x+y=10
7back into the first equation, we can solve
for x:
2x+ 3(10
7−x)−4=4
Step 6: Simplifying the equation yields:
2x+30
7−3x−4 = 4
−x=2
7
x=−2
7
Step 7: Substituting x=−2
7into x+y=10
7, we can solve for y:
−2
7+y=10
7
y=12
7
Step 8: Lastly, substituting x=−2
7and y=12
7into z= 2x+ 3y−4, we
can solve for z:
z= 2(−2
7) + 3(12
7)−4
z=−4
7+36
7−4
z=32
7−4
z=32
7−28
7
z=4
7
Step 9: Therefore, the solution to the system of equations is x=−2
7,y=12
7,
and z=4
7.
27
Question 29
Question
Solve the system of equations:
2x+ 3y−z= 6
−x+ 4y+ 2z=−10
3x+ 3y+z= 4
Solution
Step 1: Begin by writing the system of equations in matrix form:
2 3 −1
−1 4 2
331
x
y
z
=
6
−10
4
Step 2: Next, find the inverse of the coefficient matrix:
Let A=
2 3 −1
−1 4 2
331
To find A−1, use row operations to augment Awith an identity matrix and row
reduce:
2 3 −1 1 0 0
−1 4 2 0 1 0
3 3 1 0 0 1
Step 3: After row reduction, the augmented matrix becomes:
100−5
16
3
16 −1
8
010 −1
8
1
8
1
4
001 3
8−3
8
1
4
So, A−1is:
A−1=
−5
16
3
16 −1
8
−1
8
1
8
1
4
3
8−3
8
1
4
Step 4: Finally, the solution to the system is given by A−1times the constant
vector:
x
y
z
=
−5
16
3
16 −1
8
−1
8
1
8
1
4
3
8−3
8
1
4
6
−10
4
Solving for x,y, and zgives the solution to the system of equations.
28
Question 30
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 3
Solution
Step 1: Solve the second equation for y in terms of x. Step 2: Substitute the
expression for y from Step 1 into the first equation and solve for x. Step 3: Use
the value of x to find the corresponding value of y. Step 4: Verify the solution
by substituting the values of x and y back into both equations.
Step 1: Solve the second equation for yin terms of x.
4x−y= 3 =⇒y= 4x−3
Step 2: Substitute the expression for yinto the first equation and solve for
x.
2x+ 3(4x−3) = 72x+ 12x−9 = 714x−9 = 714x= 16x=16
14 =⇒x=8
7
Step 3: Use the value of xto find the corresponding value of y.
y= 4 8
7−3 = 32
7−3 = 32 −21
7=11
7
Step 4: Verify the solution by substituting the values of xand yback into
both equations.
(28
7+ 3 11
7= 7
48
7−11
7= 3 ⇒(16+33
7= 7
32−11
7= 3 ⇒(49
7= 7
21
7= 3 ⇒(7 = 7
3 = 3
Since both equations are satisfied by x=8
7and y=11
7, the solution is
correct. Thus, the solution to the system of equations is x=8
7and y=11
7.
Question 31
Question
Solve the following system of equations:
(2x+ 3y= 7
4x+ 6y= 14
29
Solution
Step 1: Let’s start by simplifying the second equation by dividing both sides
by 2:
2x+ 3y= 7 ⇒x+3
2y=7
2
Step 2: Now we have a system of equations as:
(2x+ 3y= 7
x+3
2y=7
2
Step 3: Let’s multiply the second equation by 2 to get rid of the fraction:
2x+3
2y= 2 7
2⇒2x+ 3y= 7
Step 4: We see that the first and simplified versions of the second equation
are the same. This means the two equations are actually representing the same
line, so the solution to the system of equations is any point on that line. In
other words, the system is consistent and dependent.
Question 32
Question
Solve the following system of equations:
(2x+ 3y= 8
3x−2y=−1
Solution
Step 1: Multiply the first equation by 2 and the second equation by 3 to elimi-
nate ywhen adding the two equations.
(4x+ 6y= 16
9x−6y=−3
Step 2: Add the two equations to eliminate y.
13x= 13
Step 3: Solve for xby dividing both sides by 13.
x= 1
Step 4: Substitute x= 1 back into the first equation to solve for y.
2(1) + 3y= 8
30
Step 5: Simplify and solve for y.
3y= 6 =⇒y= 2
Therefore, the solution to the system of equations is x= 1 and y= 2.
Question 33
Question
Solve the following system of equations:
3x−2y+z= 5
2x+ 3y−z= 7
x+ 4y+ 2z= 10
Solution
Step 1: Add the first and second equation to eliminate the variable z.
Step 2: Add the second and third equation to eliminate the variable z.
Step 3: Solve the resulting system of two equations for xand y.
Step 4: Substitute the values of xand yinto one of the original equations
to solve for z.
Step 1: Adding the first and second equation gives:
5x+y= 12 (Equation 4)
Step 2: Adding the second and third equation gives:
3x+ 7y= 17 (Equation 5)
Step 3: To solve Equations (4) and (5) simultaneously, we can multiply
Equation (4) by 3 and subtract it from Equation (5):
(3x+ 7y)−3(5x+y) = 17 −36
3x+ 7y−15x−3y=−19
−12x+ 4y=−19
3x−y=19
4(Equation 6)
Now we have two equations (Equations 4 and 6) that we can solve simulta-
neously for xand y.
Step 4: Substitute Equation 6 into Equation 4:
5x−3x= 19
2x= 19
31
x=19
2
Substitute xback into Equation 4:
3(19
2) + y= 12
57
2+y= 12
y= 12 −57
2
y=−33
2
Finally, substitute xand yback into one of the original equations to solve
for z. Let’s use the first equation:
3(19
2)−2(−33
2) + z= 5
57
2+66
2+z= 5
123
2+z= 5
z= 5 −123
2
z=−113
2
Therefore, the solution to the system of equations is x=19
2,y=−33
2, and
z=−113
2.
Question 34
Question
Solve the following system of equations:
(2x−3y= 5
3x+ 4y= 2
32
Solution
Step 1: Multiply the first equation by 3 and the second equation by 2 to simplify
the system:
(6x−9y= 15
6x+ 8y= 4
Step 2: Subtract the second equation from the first equation to eliminate x:
(6x−9y)−(6x+ 8y) = 15 −4
−17y= 11
y=−11
17
Step 3: Substitute the value of yback into the first equation to solve for x:
2x−3(−11
17)=5
2x+33
17 = 5
2x= 5 −33
17
2x=25
17
x=25
34
Therefore, the solution to the system of equations is:
x=25
34
y=−11
17
Question 35
Question
Solve the following system of equations using the elimination method:
(2x−3y= 7
3x+ 2y= 6
Solution
Step 1: Multiply the first equation by 3 and the second equation by 2 to elimi-
nate y:
(6x−9y= 21
6x+ 4y= 12
33
We can solve this system by substituting (x, y) = (2 + √2 cos(t),3 + √13 sin(t)),
where tis a parameter.
Thus, the solutions to the system are
(x, y) = 2 + √2 cos(t),3 + √13 sin(t)
Question 2
Question
Solve the following system of equations:
(x2+y2= 25
x+y= 7
Solution
Step 1: We recognize the second equation as the equation of a line in slope-
intercept form. We can rewrite it as y=−x+ 7.
Step 2: Next, we substitute y=−x+ 7 into the first equation x2+y2= 25:
x2+ (−x+ 7)2= 25
Step 3: Expand and simplify the expression on the left side:
x2+ (x2−14x+ 49) = 25
Step 4: Combine like terms:
2x2−14x+ 49 = 25
Step 5: Rearrange the equation into standard form:
2x2−14x+ 24 = 0
Step 6: Divide the entire equation by 2 to simplify:
x2−7x+ 12 = 0
Step 7: Factor the quadratic equation:
(x−3)(x−4) = 0
Step 8: Set each factor equal to zero and solve for x:
x−3 = 0 =⇒x= 3
x−4 = 0 =⇒x= 4
2
Step 9: Substitute x= 3 and x= 4 back into the linear equation y=−x+ 7
to find the corresponding yvalues: For x= 3:
y=−(3) + 7 = 4
For x= 4:
y=−(4) + 7 = 3
Step 10: Therefore the solution to the system of equations is:
(x, y) = {(3,4),(4,3)}
Question 3
Question
Find the solution to the following system of equations:
(x2+y2= 25
xy = 12
Solution
Step 1: We can start by squaring the second equation and simplifying to elimi-
nate y2:
(x2+y2)2= (25)2
x4+ 2x2y2+y4= 625
x4+ 2(12)2+y4= 625
x4+ 288 + y4= 625
x4+y4= 337
Step 2: Now, we substitute x2+y2= 25 and x4+y4= 337 into the equations:
(x2+y2= 25
x4+y4= 337
Step 3: Let’s denote x2as aand y2as b. The system now becomes:
(a+b= 25
a2+b2= 337
Step 4: We can solve this new system of equations to find the values of a
and b: From a+b= 25, we can express ain terms of bas a= 25 −b.
Step 5: Substitute a= 25 −binto a2+b2= 337:
(25 −b)2+b2= 337
3
625 −50b+b2+b2= 337
2b2−50b+ 288 = 0
b2−25b+ 144 = 0
(b−9)(b−16) = 0
Step 6: We find that b= 9 or b= 16. Thus, the corresponding values
for aare 16 and 9 respectively. Therefore, the possible solutions for (x, y) are
(−3,−4),(−4,−3),(3,4),and (4,3).
Question 4
Question
Find the solution to the system of equations:
(2x−3y= 5
3x+ 4y= 11
Solution
Step 1: Multiply the first equation by 3 and the second equation by 2 to make
the coefficients of xin both equations equal.
(6x−9y= 15
6x+ 8y= 22
Step 2: Subtract the first equation from the second equation to eliminate x.
17y= 7
Step 3: Solve for y.
y=7
17
Step 4: Substitute y=7
17 back into the first equation and solve for x.
2x−37
17= 5
2x−21
17 = 5
2x=104
17
x=52
17
Therefore, the solution to the system of equations is x=52
17 and y=7
17 .
4
Question 5
Question
Solve the following system of equations:
2x+ 4y−z= 5
x−3y+ 2z=−8
3x+ 2y−5z= 10
Solution
Step 1: Multiply the second equation by 2 and add it to the first equation to
eliminate x:
2x+ 4y−z= 5
4x−6y+ 4z=−16
3x+ 2y−5z= 10
Step 2: Multiply the first equation by 2 and subtract it from the third
equation to eliminate x:
2x+ 4y−z= 5
4x−6y+ 4z=−16
−5y+ 3z= 0
Step 3: Solve the third equation for yin terms of z:
−5y+ 3z= 0 =⇒y=3z
5
Step 4: Substitute y=3z
5back into the first equation:
2x+ 4 3z
5−z= 5 =⇒2x+12z
5−z= 5 =⇒2x+7z
5= 5
Step 5: Multiply the above equation by 5 to eliminate the fraction:
10x+ 7z= 25
Step 6: Substitute y=3z
5back into the second equation:
x−33z
5+ 2z=−8 =⇒x−9z
5+ 2z=−8 =⇒x−4z
5=−8
Step 7: Multiply the above equation by 5 to eliminate the fraction:
5x−4z=−40
5
Step 8: Now, we have a system of two equations:
(10x+ 7z= 25
5x−4z=−40
Step 9: Solve the above system of equations to find xand z.
(10x+ 7z)−2(5x−4z) = 25 −2(−40)
10x+ 7z−10x+ 8z= 25 + 80
15z= 105
z= 7
Step 10: Substitute z= 7 back into the equation 5x−4z=−40 to find x:
5x−4(7) = −40 =⇒5x−28 = −40 =⇒5x=−12 =⇒x=−2.4
Step 11: Substitute x=−2.4 and z= 7 back into the equation 2x+4y−z= 5
to find y:
2(−2.4) + 4y−7 = 5 =⇒ −4.8+4y−7 = 5 =⇒4y= 16.8 =⇒y= 4.2
Thus, the solution to the system of equations is:
x=−2.4, y = 4.2, z = 7
Question 6
Question
Solve the following system of equations:
(2x+ 3y= 8
4x−3y= 2
Solution
Step 1: Let’s write the system of equations:
(2x+ 3y= 8
4x−3y= 2
Step 2: Add the two equations together to eliminate y:
(2x+ 3y) + (4x−3y) = 8 + 2
6
Step 3: Simplify the equation:
6x= 10
Step 4: Solve for x:
x=10
6=5
3
Step 5: Substitute xback into one of the original equations. Let’s use the
first equation:
25
3+ 3y= 8
Step 6: Simplify the equation and solve for y:
10
3+ 3y= 8
3y= 8 −10
3
3y=24 −10
3
3y=14
3
y=14
9
Step 7: The solution to the system of equations is x=5
3and y=14
9.
Question 7
Question
Solve the following system of equations:
3x−2y+z= 7
x+ 4y−2z=−4
2x−3y+ 2z= 5
Solution
Step 1: Begin by expressing the system of equations as an augmented matrix.
3−2 1 7
1 4 −2−4
2−3 2 5
7
Step 2: Use row operations to transform the matrix to row-echelon form.
3−2 1 7
1 4 −2−4
2−3 2 5
→
1 4 −2−4
010
3−5
3
25
3
0−11 5 13
Step 3: Continue row operations to reduce the matrix to row-echelon form.
1 4 −2−4
010
3−5
3
25
3
0−11 5 13
→
1 4 −2−4
0 1 −1
2
5
2
0−11 5 13
→
1 4 −2−4
0 1 −1
2
5
2
0 0 0 0
Step 4: Interpret the row-echelon form of the system. The third equation
0 = 0 indicates that there are infinitely many solutions. The system has one
free variable, which can be denoted as t. Now express x,y, and zin terms of t.
x=−4−4t,y=5
2+1
2t,z=t
So, the solution to the system of equations is:
x=−4−4t
y=5
2+1
2t
z=t
Question 8
Question
Solve the following system of equations:
(x2+y2= 25
xy = 12
Solution
Step 1: From the second equation, we can express yas 12
x. Substituting this
into the first equation, we get:
x2+12
x2
= 25
Step 2: Simplifying the equation gives:
x2+144
x2= 25
Step 3: Multiplying through by x2gives us a quadratic equation:
x4−25x2+ 144 = 0
8
Step 4: This equation can be factored as (x2−9)(x2−16) = 0, which gives
us two sets of solutions: x=±3 and x=±4.
Step 5: We can now find the corresponding values of yby using the second
equation. For x= 3: y=12
3= 4 or y=−4 For x=−3: y=12
−3=−4 or y= 4
For x= 4: y=12
4= 3 or y=−3 For x=−4: y=12
−4=−3 or y= 3
Therefore, the solutions to the system are:
(x, y) = (3,4),(−3,−4),(4,3),(−4,−3)
Question 9
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 5
Solution
Step 1: We can start solving this system of equations by eliminating one of the
variables. Let’s eliminate y: Step 2: Multiply the second equation by 3 to make
the coefficients of yequal:
(2x+ 3y= 7
12x−3y= 15
Step 3: Add the two equations together to eliminate y:
14x= 22
Step 4: Solve for xby dividing both sides by 14:
x=22
14 =11
7
Step 5: Substitute x=11
7back into the first equation to solve for y:
211
7+ 3y= 7
22
7+ 3y= 7
Step 6: Subtract 22
7from both sides:
3y= 7 −22
7=35
7−22
7=13
7
9
Step 7: Solve for yby dividing both sides by 3:
y=13
21
Step 8: The solution to the system of equations is x=11
7and y=13
21 .
Question 10
Question
Solve the following system of equations:
(3x−2y= 5
2x2+ 5y=−1
Solution
Step 1: We can start by solving the first equation for y.
3x−2y= 5 =⇒ −2y=−3x+ 5 =⇒y=3
2x−5
2
Step 2: Substitute yin terms of xinto the second equation and solve for x.
2x2+ 5 3
2x−5
2=−1
Simplify the equation:
2x2+15
2x−25
2=−1
4x2+ 15x−25 = −2
4x2+ 15x−23 = 0
Step 3: Use the quadratic formula to find the solutions for x. The solutions
are:
x=−15 ±p152−4∗4∗(−23)
2∗4
x=−15 ±√225 + 368
8
x=−15 ±√593
8
Thus, x=−15+√593
8or x=−15−√593
8.
Step 4: Substitute each value of xback into y=3
2x−5
2to find the respective
values of y. Therefore, the solution to the system of equations is:
−15 + √593
8,7−3√593
8!and −15 −√593
8,7+3√593
8!
10
Question 11
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 5
Solution
Step 1: Solve the second equation for yin terms of x.
4x−y= 5
y= 4x−5
Step 2: Substitute yfrom the second equation into the first equation.
2x+ 3(4x−5) = 7
2x+ 12x−15 = 7
14x−15 = 7
Step 3: Solve for x.
14x= 22
x=22
14
x=11
7
Step 4: Substitute the value of xback into the equation y= 4x−5 to find
y.
y= 4 11
7−5
y=44
7−5
y=44
7−35
7
y=9
7
Step 5: The solution to the system of equations is x=11
7and y=9
7.
Question 12
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 5
11
Solution
Step 1: Solve the second equation for y.
4x−y= 5
y= 4x−5
Step 2: Substitute y= 4x−5 into the first equation and solve for x.
2x+ 3(4x−5) = 7
2x+ 12x−15 = 7
14x−15 = 7
14x= 22
x=22
14
x=11
7
Step 3: Substitute x=11
7into y= 4x−5 and solve for y.
y= 4 11
7−5
y=44
7−5
y=44
7−35
7
y=9
7
Therefore, the solution to the system of equations is x=11
7and y=9
7.
Question 13
Question
Solve the following system of equations:
(2x+ 3y= 7
3x−2y= 8
Solution
Step 1: Solve the first equation for x.
2x+ 3y= 7 =⇒x=7−3y
2
12
Step 2: Substitute the expression for xinto the second equation.
37−3y
2−2y= 8
Step 3: Simplify the equation.
21 −9y
2−2y= 8
Step 4: Multiply through by 2 to clear the fractions.
21 −9y−4y= 16
Step 5: Combine like terms.
21 −13y= 16
Step 6: Solve for y.
−13y=−5 =⇒y=5
13
Step 7: Substitute the value of yback into the equation to find x.
x=7−35
13
2
Step 8: Simplify to find the value of x.
x=7−15
13
2=7·13 −15
2·13 =91 −15
26 =76
26 =38
13
Therefore, the solution to the system of equations is x=38
13 and y=5
13 .
Question 14
Question
Solve the following system of equations:
(x2−2xy +y2= 25
x2+ 2xy +y2= 49
Solution
Step 1: Start by subtracting the second equation from the first equation to
eliminate 2xy.
x2−2xy +y2= 25
x2+ 2xy +y2= 49
−4xy =−24
xy = 6
13
Step 2: Next, we can rewrite the original first equation in terms of xy.
(x+y)2=x2+ 2xy +y2= 49
x+y=√49
x+y= 7 (since x, y are positive)
Step 3: Now we have a system of equations:
(x+y= 7
xy = 6
We can solve this system by substituting y= 7 −xinto the second equation.
x(7 −x) = 6
7x−x2= 6
x2−7x+ 6 = 0
(x−1)(x−6) = 0
Step 4: From here, we have two possible solutions:
x= 1, y = 6
x= 6, y = 1
Therefore, the solutions to the system of equations are x= 1, y = 6 and
x= 6, y = 1.
Question 15
Question
Solve the following system of equations:
(2x+ 3y= 11
3x−2y= 1
Solution
Step 1: We will solve the system of equations using the method of substitution.
We can rearrange the first equation to solve for xin terms of y:
2x+ 3y= 11 ⇒2x= 11 −3y⇒x=11 −3y
2
Step 2: Substitute the expression for xinto the second equation:
311 −3y
2−2y= 1
14
Step 3: Simplify the equation and solve for y:
33 −9y
2−2y= 1 ⇒33−9y−4y= 2 ⇒33−13y= 2 ⇒ −13y=−31 ⇒y=31
13 =31
13
Step 4: Substitute yback into the expression for xto solve for x:
x=11 −3·31
13
2=11 −93/13
2=11 ·13 −93
13 ·2=143 −93
26 =50
26 =25
13
Step 5: The solution to the system of equations is x=25
13 and y=31
13 .
Question 16
Question
Solve the following system of equations for xand y:
(3x+ 2y= 7
5x−4y= 1
Solution
Step 1: Multiply the first equation by 2 and the second equation by 3 to elimi-
nate y:
(6x+ 4y= 14
15x−12y= 3
Step 2: Add the two equations together to eliminate y:
21x= 17
Step 3: Solve for x:
x=17
21
Step 4: Substitute xback into the first equation to solve for y:
317
21+ 2y= 7
17 + 2y= 7
2y=−10
y=−5
Therefore, the solution to the system of equations is x=17
21 and y=−5.
15
Question 17
Question
Solve the following system of equations:
(3x−2y= 4
x2+y2= 25
Solution
Step 1: Solve the first equation for xin terms of y.
3x−2y= 4
3x= 2y+ 4
x=2y+ 4
3
Step 2: Substitute xinto the second equation.
2y+ 4
32
+y2= 25
4y2+ 16y+ 16
9+y2= 25
4y2+ 16y+ 16 + 9y2= 225
13y2+ 16y−209 = 0
Step 3: Solve the quadratic equation for y.
y=−b±√b2−4ac
2a
y=−16 ±p162−4(13)(−209)
2(13)
y=−16 ±√256 + 10868
26
y=−16 ±√11124
26
y=−16 ±106
26 (since √11124 = 106)
y=90
26 or y=−122
26
y=45
13 or y=−61
13
16
Step 4: Substitute the yvalues back to solve for x.
x=2(45
13 )+4
3=94
13
x=2(−61
13 )+4
3=−34
13
Therefore, the solutions to the system of equations are x=94
13 ,y=45
13 and
x=−34
13 ,y=−61
13 .
Question 18
Question
Solve the following system of equations:
(3x−2y= 4
2x+y= 1
Solution
Step 1: We can solve this system of equations using either substitution or elim-
ination method. Let’s use the elimination method.
Step 2: Multiply the second equation by 2:
(3x−2y= 4
4x+ 2y= 2
Step 3: Add the two equations together to eliminate y:
7x= 6
Step 4: Solve for x:
x=6
7
Step 5: Substitute the value of xback into the second equation and solve
for y:
26
7+y= 1
y= 1 −12
7
y=−5
7
Step 6: Therefore, the solution to the given system of equations is:
x=6
7, y =−5
7
17
Question 19
Question
Solve the following system of equations:
(2x+ 3y= 10
4x−y= 5
Solution
Step 1: Solve the second equation for yin terms of x.
4x−y= 5
y= 4x−5
Step 2: Substitute y= 4x−5 into the first equation and solve for x.
2x+ 3(4x−5) = 10
2x+ 12x−15 = 10
14x−15 = 10
14x= 25
x=25
14
Step 3: Substitute x=25
14 back into y= 4x−5 and solve for y.
y= 4 25
14−5
y=100
14 −70
14
y=30
14
y=15
7
Therefore, the solution to the system of equations is x=25
14 and y=15
7.
Question 20
Question
Find the solution to the following system of equations:
(3x+ 4y= 5
2x−y= 8
18
Solution
Step 1: Let’s rewrite the system of equations in standard form:
3x+ 4y= 5
2x−y= 8
Step 2: We will use the elimination method to solve the system. To eliminate
y, we multiply equation (2) by 4 and add it to equation (1):
3x+ 4y= 5
8x−4y= 32
Step 3: Adding the two equations we get:
11x= 37
Step 4: Solve for x:
x=37
11
Step 5: Substitute xback into equation (2) to solve for y:
237
11−y= 8
Step 6: Simplify and solve for y:
y=1
11
Step 7: Therefore, the solution to the system of equations is x=37
11 and
y=1
11 .
Question 21
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 5
Solution
Step 1: Solve the second equation for y.
4x−y= 5
⇒y= 4x−5
19
Step 2: Substitute y= 4x−5 into the first equation.
2x+ 3(4x−5) = 7
2x+ 12x−15 = 7
14x−15 = 7
14x= 22
x=22
14
x=11
7
Step 3: Substitute x=11
7into y= 4x−5 to solve for y.
y= 4 11
7−5
y=44
7−5
y=44
7−35
7
y=9
7
So, the solution to the system of equations is x=11
7and y=9
7.
Question 22
Question
Consider the following system of equations:
(3x−y= 5
x2−2xy +y2= 9
Solve the system of equations.
Solution
Step 1: We will start by solving the first equation for yin terms of x.
3x−y= 5
y= 3x−5
20
Step 2: Substituting y= 3x−5 into the second equation, we get:
x2−2x(3x−5) + (3x−5)2= 9
x2−6x2+ 10x+ 9x2−30x+ 25 = 9
−2x2−20x+ 25 = 9
−2x2−20x+ 16 = 0
x2+ 10x−8 = 0
Step 3: Next, we’ll solve the quadratic equation x2+ 10x−8 = 0 using the
quadratic formula:
x=−b±√b2−4ac
2a
x=−10 ±p102−4(1)(−8)
2(1)
x=−10 ±√100 + 32
2
x=−10 ±√132
2
x=−10 ±2√33
2
x=−5±√33
So, the solutions for xare x1=−5 + √33 and x2=−5−√33.
Step 4: Using the values of x1and x2to find the corresponding yvalues:
y1= 3(−5 + √33) −5 = −15 + 3√33 −5 = −20 + 3√33
y2= 3(−5−√33) −5 = −15 −3√33 −5 = −20 −3√33
Therefore, the solutions to the system of equations are (−5 + √33,−20 +
3√33) and (−5−√33,−20 −3√33).
Question 23
Question
Find the solution to the system of equations:
(2x+ 3y= 7
4x−3y= 5
21
Solution
Step 1: Add the two equations together to eliminate y:
(2x+ 3y) + (4x−3y) = 7 + 5
6x= 12
x= 2
Step 2: Substitute x= 2 back into the first equation to solve for y:
2(2) + 3y= 7
4+3y= 7
3y= 3
y= 1
Step 3: The solution to the system of equations is x= 2 and y= 1.
Question 24
Question
Solve the following system of equations:
3x−y+z= 2
2x+y−2z= 3
x+ 3y−4z=−1
Solution
Step 1: We will use the method of substitution to solve this system of equations.
Let’s start by solving the first equation for y: From the first equation:
y= 3x+z−2
Step 2: Next, we substitute y= 3x+z−2 into the second and third equations
to eliminate y: Substitute y= 3x+z−2 into the second equation:
2x+ (3x+z−2) −2z= 3
Simplify to get:
5x+z−2z= 5
5x−z= 5 (Equation 4)
Step 3: Substitute y= 3x+z−2 into the third equation:
x+ 3(3x+z−2) −4z=−1
22
Simplify to get:
x+ 9x+ 3z−6−4z=−1
10x−z= 5 (Equation 5)
Step 4: Now, we will solve Equations 4 and 5 simultaneously. Multiply
Equation 4 by 2:
2(5x−z) = 2(5)
10x−2z= 10 (Equation 6)
Step 5: Subtract Equation 5 from Equation 6:
(10x−2z)−(10x−z) = 10 −5
−x= 5
x=−5
Step 6: Substitute x=−5 into Equation 4:
5(−5) −z= 5
−25 −z= 5
z=−30
Step 7: Substitute x=−5 and z=−30 into Equation 1 to solve for y:
3(−5) −y+ (−30) = 2
−15 −y−30 = 2
y=−13
Therefore, the solution to the system of equations is:
x=−5
y=−13
z=−30
Question 25
Question
Solve the system of equations:
(4x−3y= 5
2x2+ 3xy = 1
23
Solution
Step 1: Let’s start by multiplying the first equation by 2 to eliminate ywhen
adding the equations. Step 2: We will then substitute the value of yback into
the first equation to solve for x. Step 3: Finally, we will substitute the value of
xback into the first equation to solve for y.
Step 1: Multiply the first equation by 2:
2(4x−3y) = 2(5)
8x−6y= 10
Step 2: Substitute 8x−6yinto the second equation:
2x2+ 3x(8x−6y)=1
2x2+ 24x2−18xy = 1
26x2−18xy = 1
Step 3: Substitute yback into the first equation:
4x−326x2−1
18x= 5
4x−78x2−3
18 = 5
72x2−54x−78x2+ 3 = 90
−6x2−54x+ 3 = 90
−6x2−54x−87 = 0
2x2+ 18x+ 29 = 0
This is a quadratic equation, so we can apply the quadratic formula to solve
for x:
x=−18 ±p182−4(2)(29)
2(2)
x=−18 ±√324 −232
4
x=−18 ±√92
4
x=−18 ±2√23
4
x=−9
2±√23
2
Therefore, the solutions for the system of equations are:
x=−9
2+√23
2
y=−5
3−5√23
18
24
Question 26
Question
Solve the following system of equations:
(3x+ 2y= 5
6x+ 3y= 8
Solution
Step 1: Multiply the first equation by 2 to make the coefficients of ythe same:
(6x+ 4y= 10
6x+ 3y= 8
Step 2: Subtract the second equation from the first equation:
(6x+ 4y)−(6x+ 3y) = 10 −8
y= 2
Step 3: Substitute y= 2 back into the first equation to solve for x:
3x+ 2(2) = 5
3x+ 4 = 5
3x= 1
x=1
3
So the solution to the system of equations is x=1
3and y= 2.
Question 27
Question
Solve the following system of equations:
(2x+ 3y= 7
x2+y2= 10
Solution
To solve this system of equations, we will first solve for one variable in terms of
the other and then substitute back to find the values of both variables.
25
Step 1: Solve for xin terms of yfrom the first equation:
2x+ 3y= 7
2x= 7 −3y
x=7
2−3
2y
Step 2: Substitute xinto the second equation:
7
2−3
2y2
+y2= 10
49
4−21
2y+9
4y2+y2= 10
49
4−21
2y+13
4y2= 10
52 −42y+ 13y2= 40
13y2−42y+ 12 = 0
Step 3: Solve the quadratic equation: Solving the quadratic equation
using the quadratic formula:
y=42 ±p422−4(13)(12)
2(13)
y=42 ±√1764 −624
26
y=42 ±√1140
26
y=42 ±2√285
26
y=21 ±√285
13
Therefore, the solutions to the system of equations are:
(x=7
2−3
221±√285
13
y=21±√285
13
Question 28
Question
Find all solutions to the following system of equations:
2x+ 3y−z= 4
x−2y+ 3z=−2
3x+ 2y+ 4z= 10
26
Solution
Step 1: We will solve the system using the method of substitution.
Step 2: From the first equation, we can isolate z:
z= 2x+ 3y−4
Step 3: Substituting zin the second equation, we get:
x−2y+ 3(2x+ 3y−4) = −2
Step 4: Simplifying the equation, we obtain:
x−2y+ 6x+ 9y−12 = −2
7x+ 7y= 10
x+y=10
7
Step 5: Substituting x+y=10
7back into the first equation, we can solve
for x:
2x+ 3(10
7−x)−4=4
Step 6: Simplifying the equation yields:
2x+30
7−3x−4 = 4
−x=2
7
x=−2
7
Step 7: Substituting x=−2
7into x+y=10
7, we can solve for y:
−2
7+y=10
7
y=12
7
Step 8: Lastly, substituting x=−2
7and y=12
7into z= 2x+ 3y−4, we
can solve for z:
z= 2(−2
7) + 3(12
7)−4
z=−4
7+36
7−4
z=32
7−4
z=32
7−28
7
z=4
7
Step 9: Therefore, the solution to the system of equations is x=−2
7,y=12
7,
and z=4
7.
27
Question 29
Question
Solve the system of equations:
2x+ 3y−z= 6
−x+ 4y+ 2z=−10
3x+ 3y+z= 4
Solution
Step 1: Begin by writing the system of equations in matrix form:
2 3 −1
−1 4 2
331
x
y
z
=
6
−10
4
Step 2: Next, find the inverse of the coefficient matrix:
Let A=
2 3 −1
−1 4 2
331
To find A−1, use row operations to augment Awith an identity matrix and row
reduce:
2 3 −1 1 0 0
−1 4 2 0 1 0
3 3 1 0 0 1
Step 3: After row reduction, the augmented matrix becomes:
100−5
16
3
16 −1
8
010 −1
8
1
8
1
4
001 3
8−3
8
1
4
So, A−1is:
A−1=
−5
16
3
16 −1
8
−1
8
1
8
1
4
3
8−3
8
1
4
Step 4: Finally, the solution to the system is given by A−1times the constant
vector:
x
y
z
=
−5
16
3
16 −1
8
−1
8
1
8
1
4
3
8−3
8
1
4
6
−10
4
Solving for x,y, and zgives the solution to the system of equations.
28
Question 30
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 3
Solution
Step 1: Solve the second equation for y in terms of x. Step 2: Substitute the
expression for y from Step 1 into the first equation and solve for x. Step 3: Use
the value of x to find the corresponding value of y. Step 4: Verify the solution
by substituting the values of x and y back into both equations.
Step 1: Solve the second equation for yin terms of x.
4x−y= 3 =⇒y= 4x−3
Step 2: Substitute the expression for yinto the first equation and solve for
x.
2x+ 3(4x−3) = 72x+ 12x−9 = 714x−9 = 714x= 16x=16
14 =⇒x=8
7
Step 3: Use the value of xto find the corresponding value of y.
y= 4 8
7−3 = 32
7−3 = 32 −21
7=11
7
Step 4: Verify the solution by substituting the values of xand yback into
both equations.
(28
7+ 3 11
7= 7
48
7−11
7= 3 ⇒(16+33
7= 7
32−11
7= 3 ⇒(49
7= 7
21
7= 3 ⇒(7 = 7
3 = 3
Since both equations are satisfied by x=8
7and y=11
7, the solution is
correct. Thus, the solution to the system of equations is x=8
7and y=11
7.
Question 31
Question
Solve the following system of equations:
(2x+ 3y= 7
4x+ 6y= 14
29
Solution
Step 1: Let’s start by simplifying the second equation by dividing both sides
by 2:
2x+ 3y= 7 ⇒x+3
2y=7
2
Step 2: Now we have a system of equations as:
(2x+ 3y= 7
x+3
2y=7
2
Step 3: Let’s multiply the second equation by 2 to get rid of the fraction:
2x+3
2y= 2 7
2⇒2x+ 3y= 7
Step 4: We see that the first and simplified versions of the second equation
are the same. This means the two equations are actually representing the same
line, so the solution to the system of equations is any point on that line. In
other words, the system is consistent and dependent.
Question 32
Question
Solve the following system of equations:
(2x+ 3y= 8
3x−2y=−1
Solution
Step 1: Multiply the first equation by 2 and the second equation by 3 to elimi-
nate ywhen adding the two equations.
(4x+ 6y= 16
9x−6y=−3
Step 2: Add the two equations to eliminate y.
13x= 13
Step 3: Solve for xby dividing both sides by 13.
x= 1
Step 4: Substitute x= 1 back into the first equation to solve for y.
2(1) + 3y= 8
30
Step 5: Simplify and solve for y.
3y= 6 =⇒y= 2
Therefore, the solution to the system of equations is x= 1 and y= 2.
Question 33
Question
Solve the following system of equations:
3x−2y+z= 5
2x+ 3y−z= 7
x+ 4y+ 2z= 10
Solution
Step 1: Add the first and second equation to eliminate the variable z.
Step 2: Add the second and third equation to eliminate the variable z.
Step 3: Solve the resulting system of two equations for xand y.
Step 4: Substitute the values of xand yinto one of the original equations
to solve for z.
Step 1: Adding the first and second equation gives:
5x+y= 12 (Equation 4)
Step 2: Adding the second and third equation gives:
3x+ 7y= 17 (Equation 5)
Step 3: To solve Equations (4) and (5) simultaneously, we can multiply
Equation (4) by 3 and subtract it from Equation (5):
(3x+ 7y)−3(5x+y) = 17 −36
3x+ 7y−15x−3y=−19
−12x+ 4y=−19
3x−y=19
4(Equation 6)
Now we have two equations (Equations 4 and 6) that we can solve simulta-
neously for xand y.
Step 4: Substitute Equation 6 into Equation 4:
5x−3x= 19
2x= 19
31
x=19
2
Substitute xback into Equation 4:
3(19
2) + y= 12
57
2+y= 12
y= 12 −57
2
y=−33
2
Finally, substitute xand yback into one of the original equations to solve
for z. Let’s use the first equation:
3(19
2)−2(−33
2) + z= 5
57
2+66
2+z= 5
123
2+z= 5
z= 5 −123
2
z=−113
2
Therefore, the solution to the system of equations is x=19
2,y=−33
2, and
z=−113
2.
Question 34
Question
Solve the following system of equations:
(2x−3y= 5
3x+ 4y= 2
32
Solution
Step 1: Multiply the first equation by 3 and the second equation by 2 to simplify
the system:
(6x−9y= 15
6x+ 8y= 4
Step 2: Subtract the second equation from the first equation to eliminate x:
(6x−9y)−(6x+ 8y) = 15 −4
−17y= 11
y=−11
17
Step 3: Substitute the value of yback into the first equation to solve for x:
2x−3(−11
17)=5
2x+33
17 = 5
2x= 5 −33
17
2x=25
17
x=25
34
Therefore, the solution to the system of equations is:
x=25
34
y=−11
17
Question 35
Question
Solve the following system of equations using the elimination method:
(2x−3y= 7
3x+ 2y= 6
Solution
Step 1: Multiply the first equation by 3 and the second equation by 2 to elimi-
nate y:
(6x−9y= 21
6x+ 4y= 12
33
Step 2: Subtract the second equation from the first equation to eliminate x:
(6x−9y)−(6x+ 4y) = 21 −12
−13y= 9
y=−9
13
Step 3: Substitute y=−9
13 back into the first equation to solve for x:
2x−3(−9
13)=7
2x+27
13 = 7
2x= 7 −27
13
2x=91
13 −27
13
2x=64
13
x=32
13
Therefore, the solution to the system of equations is x=32
13 and y=−9
13 .
34
