MATH 114 - QUANTITATIVE
REASONING - Systems of Equations
Question Bank - Set 4
Liberty University
Question 1
Question
Solve the following system of equations:
(2x−3y= 5
4x+y= 1
Solution
Step 1: Let’s solve the first equation for x:
2x−3y= 5 =⇒2x= 3y+ 5 =⇒x=3y+ 5
2
Step 2: Substitute the expression for xinto the second equation:
43y+ 5
2+y= 1
Step 3: Now, solve for y:
6y+ 10 + y= 2 =⇒7y+ 10 = 2 =⇒7y=−8 =⇒y=−8
7
Step 4: Substitute yback into the equation we found for xin Step 1:
x=3(−8
7)+5
2=−24/7 + 35/7
2=11/7
2=11
14
Step 5: Therefore, the solution to the system of equations is:
(x=11
14
y=−8
7
Question 2
Question
Solve the following system of equations:
(2x−3y= 7
4x+ 5y= 13
Solution
Step 1: Multiply the first equation by 2 to eliminate xwhen adding the equa-
tions. (4x−6y= 14
4x+ 5y= 13
Step 2: Subtract the first equation from the second to eliminate x.
11y=−1
Step 3: Solve for y.
y=−1
11
Step 4: Substitute yback into one of the original equations to solve for x.
2x−3(−1
11)=7
2x+3
11 = 7
2x=76
11
x=38
11
Step 5: The solution to the system is:
(x=38
11
y=−1
11
Question 3
Question
Solve the following system of equations:
(3x+ 2y= 10
x2+y2= 25
2
Solution
Step 1: We can start by isolating xin the first equation:
3x+ 2y= 10 =⇒3x= 10 −2y=⇒x=10 −2y
3
Step 2: Substitute the expression for xinto the second equation:
10 −2y
32
+y2= 25
Step 3: Simplifying the equation by expanding and combining like terms, we
get:
(10 −2y)2
9+y2= 25
Step 4: Further simplifying the equation by expanding and combining like
terms, we get:
100 −40y+ 4y2
9+y2= 25 =⇒100 −40y+ 4y2+ 9y2= 225
Step 5: Rearranging the terms, we have a quadratic equation:
13y2−40y−125 = 0
Step 6: Solve the quadratic equation for y. We can factor or use the quadratic
formula to find the solutions.
Step 7: Once we find the values for y, we can substitute them back into the
equation we found for x. This will give us the corresponding values for x.
Step 8: Finally, we check the solutions by substituting them into both orig-
inal equations to ensure they satisfy both equations.
Question 4
Question
Solve the following system of equations:
(2x+y= 5
3x−2y= 8
Solution
Step 1: We will solve the first equation for yin terms of x.
2x+y= 5
y= 5 −2x
3
Step 2: We will substitute the expression for yinto the second equation and
solve for x.3x−2(5 −2x)=8
3x−10 + 4x= 8
7x−10 = 8
7x= 18
x=18
7
Step 3: Now, we will substitute the value of xback into the first equation
to find y.
218
7+y= 5
36
7+y= 5
y= 5 −36
7
y=35
7−36
7
y=−1
7
Therefore, the solution to the system of equations is x=18
7and y=−1
7.
Question 5
Question
Solve the following system of equations:
(3x−2y= 5
2x+y= 3
Solution
Step 1: Start by solving the second equation for y.
2x+y= 3
y= 3 −2x
Step 2: Substitute this expression for yinto the first equation.
3x−2(3 −2x)=5
3x−6+4x= 5
7x= 11
x=11
7
4
Step 3: Substitute the value of xback into the equation y= 3 −2xto find
y.
y= 3 −211
7
y= 3 −22
7
y=1
7
So, the solution to the system of equations is x=11
7and y=1
7.
Question 6
Question
Solve the following system of equations:
2x−3y+z= 7
x+ 2y−2z=−1
3x−y+z= 5
Solution
Step 1: Write the system of equations in matrix form and represent it as an
augmented matrix.
2−3 1 7
1 2 −2−1
3−1 1 5
Step 2: Perform row operations to obtain row-echelon form.
1 2 −2−1
0−7 5 8
0−7 5 8
Step 3: Subtract the second row from the third row to get a new row.
1 2 −2−1
0−7 5 8
0 0 0 0
Step 4: Express the system of equations in its simplified form.
x+ 2y−2z=−1
−7y+ 5z= 8
0 = 0
5
Step 5: Solve the second equation for y.
−7y+ 5z= 8 =⇒y=5z−8
7
Step 6: Substitute the expression for yinto the first equation and solve for
x.
x+ 2 5z−8
7−2z=−1
Simplify the equation and express xin terms of z.
Step 7: Substitute the expressions for xand yback into one of the original
equations to solve for z.
Step 8: Once zis found, substitute back into the expressions for xand y
to obtain their values.
Question 7
Question
Solve the following system of equations:
3x+ 2y−z= 7
x−y+z=−2
2x+ 3y+z= 6
Solution
Step 1: Begin by writing the system of equations as an augmented matrix:
3 2 −1 7
1−1 1 −2
2 3 1 6
Step 2: Perform row operations to simplify the matrix.
3 2 −1 7
1−1 1 −2
2 3 1 6
→
1−1 1 −2
3 2 −1 7
2 3 1 6
Step 3: Use row operations to simplify the matrix further.
1−1 1 −2
3 2 −1 7
2 3 1 6
→
1−1 1 −2
0 5 −4 13
0 5 −1 10
Step 4: Continue to simplify the matrix by performing row operations.
1−1 1 −2
0 5 −4 13
0 5 −1 10
→
1−1 1 −2
0 5 −4 13
0 0 3 −3
6
Step 5: Use the row echelon form of the matrix to back-substitute and find
the values of the variables. From the last row, we find that 3z=−3⇒z=−1.
Substituting z=−1 into the middle row equation, we get 5y+ 4 = 13 ⇒y= 3.
Finally, substituting y= 3 and z=−1 into the top row equation, we get
x−3−1 = −2⇒x= 0.
Therefore, the solution to the system of equations is x= 0, y= 3, z=−1.
Question 8
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−5y=−3
Solution
Step 1: Solve the first equation for xin terms of y.
2x+ 3y= 7 ⇒2x= 7 −3y⇒x=7−3y
2
Step 2: Substitute xin the second equation.
47−3y
2−5y=−3
Step 3: Simplify and solve for y.
14 −6y−5y=−3⇒14 −11y=−3⇒ −11y=−17 ⇒y=17
11
Step 4: Substitute yback to find x.
x=7−3(17
11 )
2
Step 5: Calculate the value of x.
x=7−51
11
2=77 −51
22 =26
22 =13
11
Step 6: The solution to the system of equations is x=13
11 and y=17
11 .
Question 9
Question
Solve the following system of equations:
(2x+ 3y= 8
3x−4y= 5
7
Solution
Step 1: Let’s start by multiplying the first equation by 3 and the second equation
by 2 to eliminate xwhen we add the equations together.
(6x+ 9y= 24
6x−8y= 10
Step 2: Now, let’s subtract the second equation from the first equation to
eliminate x.
17y= 14
Step 3: Solve for y.
y=14
17
Step 4: Substitute y=14
17 back into one of the original equations to solve
for x. Let’s use the first equation.
2x+ 3 14
17= 8
Step 5: Simplify the equation and solve for x.
2x+42
17 = 8
Step 6: Solve for x.
x=70
17
Therefore, the solution to the system of equations is x=70
17 and y=14
17 .
Question 10
Question
Solve the following system of equations:
2x−3y+z= 7
3x+ 2y−2z=−8
x+y+z= 1
Solution
Step 1: We will use the method of substitution to solve this system. Let’s start
by isolating one variable in the third equation. Subtracting yand zfrom both
sides of the third equation, we get:
x= 1 −y−z
8
Step 2: Now substitute x= 1 −y−zinto the first and second equations to
create two new equations with yand zas variables. Substitute xinto the first
equation:
2(1 −y−z)−3y+z= 7
2−2y−2z−3y+z= 7
2−5y−z= 7
5y+z=−5
This is our new equation 1.
Step 3: Substitute x= 1 −y−zinto the second equation:
3(1 −y−z)+2y−2z=−8
3−3y−3z+ 2y−2z=−8
3−y−5z=−8
y+ 5z= 11
This is our new equation 2.
Step 4: Now we have a system of two equations in two variables yand z:
(5y+z=−5
y+ 5z= 11
Step 5: We can now solve this new system of equations by multiplying the
first equation by 5 and adding the equations to eliminate y:
25y+ 5z=−25
5y+z=−5
Adding the two equations gives:
30y=−30
y=−1
Step 6: Finally, substitute y=−1 back into equation 1:
5(−1) + z=−5
−5 + z=−5
z= 0
Step 7: Now that we have y=−1 and z= 0, substitute these values into
x= 1 −y−zto find x:
x= 1 −(−1) −0
9
x= 2
Step 8: Therefore, the solution to the system of equations is:
x= 2
y=−1
z= 0
Question 11
Question
Solve the following system of equations:
(2x+ 3y= 7
4x+ 5y= 11
Solution
Step 1: Let’s start by multiplying the first equation by 2 to match the coeffi-
cients of x in both equations:
(4x+ 6y= 14
4x+ 5y= 11
Step 2: Now, subtract the second equation from the first equation to elim-
inate x: (4x+ 6y)−(4x+ 5y) = 14 −11
6y−5y= 3
y= 3
Step 3: Substitute the value of y back into one of the original equations to
solve for x. Let’s use the first equation:
2x+ 3(3) = 7
Step 4: Solve for x:
2x+ 9 = 7
2x=−2
x=−1
Step 5: Therefore, the solution to the system of equations is x=−1 and
y= 3.
10
Question 12
Question
Solve the following system of equations:
(3x+ 2y= 7
4x−y= 5
Solution
Step 1: We will solve the second equation for y:
y= 4x−5
Step 2: Substitute the expression for yinto the first equation:
3x+ 2(4x−5) = 7
Step 3: Simplify the equation:
3x+ 8x−10 = 7
11x−10 = 7
Step 4: Solve for x:
11x= 17
x=17
11
Step 5: Substitute the value of xback into the equation for y:
y= 4 17
11−5
y=68
11 −55
11
y=13
11
Step 6: Therefore, the solution to the system of equations is:
x=17
11, y =13
11
Question 13
Question
Solve the following system of equations:
3x+ 2y+z= 9
4x−y+ 2z= 7
x+ 3y−z= 4
11
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 z:
3x+ 2y+z= 9
z= 9 −3x−2y
Step 2: Now, substitute z= 9 −3x−2yinto the second equation:
4x−y+ 2(9 −3x−2y)=7
4x−y+ 18 −6x−4y= 7
−2x−5y=−11
2x+ 5y= 11 (⋆)
Step 3: Next, substitute z= 9 −3x−2yinto the third equation:
x+ 3y−(9 −3x−2y)=4
x+ 3y−9+3x+ 2y= 4
4x+ 5y= 13 (⋆⋆)
Step 4: Now, we have the system of equations:
(2x+ 5y= 11
4x+ 5y= 13
Step 5: Subtract the first equation from the second equation to get:
(4x+ 5y)−(2x+ 5y) = 13 −11
2x= 2
x= 1
Step 6: Substitute x= 1 into equation (⋆):
2(1) + 5y= 11
2+5y= 11
5y= 9
y=9
5
Step 7: Substitute x= 1 and y=9
5into one of the original equations (e.g.
the first equation) to solve for z:
3(1) + 2 9
5+z= 9
3 + 18
5+z= 9
z= 9 −3−18
5
z=12
5
12
Therefore, the solution to the system of equations is x= 1, y =9
5, z =12
5.
Question 14
Question
Solve the following system of equations:
(2x+ 3y= 5
4x−5y= 7
Solution
Step 1: Multiply the first equation by 2 to eliminate xwhen adding the two
equations. This gives us
(4x+ 6y= 10
4x−5y= 7
Step 2: Subtract the second equation from the first equation to eliminate x:
(4x+ 6y)−(4x−5y) = 10 −7
11y= 3
y=3
11
Step 3: Substitute the value of yback into the first equation to solve for x:
2x+ 3 3
11= 5
2x+9
11 = 5
2x= 5 −9
11
2x=55
11 −9
11
2x=46
11
x=46
22
x=23
11
Therefore, the solution to the system of equations is x=23
11 and y=3
11 .
13
Question 15
Question
Solve the following system of equations:
(2x+ 3y= 11
3x−2y= 4
Solution
Step 1: Start by multiplying the first equation by 3 and the second equation by
2 to get the coefficients of xto be the same:
(6x+ 9y= 33
6x−4y= 8
Step 2: Now, subtract the second equation from the first equation:
13y= 25
Step 3: Solve for y:
y=25
13
Step 4: Substitute yback into the first equation to solve for x:
2x+ 3 25
13= 11
Step 5: Simplify the equation and solve for x:
2x+75
13 = 11
Step 6:
2x= 11 −75
13 =143
13 −75
13 =68
13
Step 7:
x=68
13∇ · 2 = 68
13 ×1
2=34
13
So, the solution to the system of equations is x=34
13 and y=25
13 .
Question 16
Question
Solve the following system of equations:
(5x+ 3y= 16
3x−2y= 4
14
Solution
Step 1: Multiply the first equation by 2 and the second equation by 3 to create
opposite coefficients for yand eliminate ywhen combining the equations.
(10x+ 6y= 32
9x−6y= 12
Step 2: Add the new equations together to eliminate y.
(10x+ 6y) + (9x−6y) = 32 + 12
19x= 44
x=44
19
Step 3: Substitute xback into one of the original equations to solve for y.
Using the first equation:
544
19+ 3y= 16
220
19 + 3y= 16
3y= 16 −220
19
3y=224
19
y=224
57
Therefore, the solution to the system of equations is x=44
19 and y=224
57 .
Question 17
Question
Solve the following system of equations:
(2x+ 3y= 5
3x−y= 7
Solution
Step 1: Solve the second equation for yin terms of x.
3x−y= 7
y= 3x−7
15
Step 2: Substitute the expression for yinto the first equation.
2x+ 3(3x−7) = 5
2x+ 9x−21 = 5
11x= 26
x=26
11 =2
1
Step 3: Substitute the value of xback into the second equation to solve for
y.
3(2
1)−y= 7
6−y= 7
−y= 1
y=−1
Therefore, the solution to the system of equations is x=2
1and y=−1.
Question 18
Question
Solve the following system of equations:
(2x−3y= 5
x2+y2= 25
Solution
Step 1: Multiply the first equation by 2 to simplify the system:
(4x−6y= 10
x2+y2= 25
Step 2: Square the first equation to eliminate y:
((4x−6y)2= 100
x2+y2= 25
Step 3: Expand and simplify the squared equation:
(16x2−48xy + 36y2= 100
x2+y2= 25
16
Step 4: Substitute y2= 25 −x2from the second equation into the first
equation:
(16x2−48x(5 −x) + 36(25 −x2) = 100
x2+ 25 −x2= 25
Step 5: Simplify the system of equations:
(16x2−240x+ 900 −36x2= 100
25 = 25
Step 6: Combine like terms and solve for x:
(−20x+ 800 = 100
25 = 25
Step 7: Solve for x:
(−20x=−700
25 = 25
Step 8: Find the value of x:
x=−700
−20 = 35
Step 9: Substitute x= 35 back into x2+y2= 25 to solve for y:
352+y2= 25
1225 + y2= 25
y2=−1200
Step 10: There are no real solutions for ysince y2cannot be negative.
Therefore, the system has no real solutions.
Therefore, the system of equations has no real solutions.
Question 19
Question
Solve the following system of equations:
2x−3y+ 5z= 4
x+y−z= 3
4x−5y+z= 7
Solution
Step-by-step solution goes here.
17
Question 20
Question
Solve the following system of equations:
(3x−2y= 7
2x+y= 4
Solution
To solve this system of equations, we can use the method of substitution or
elimination.
Step 1: Let’s use the method of substitution. Solve the second equation for
yin terms of x:
2x+y= 4 =⇒y= 4 −2x
Step 2: Substitute y= 4 −2xinto the first equation:
3x−2(4 −2x) = 7
Step 3: Expand and simplify the equation:
3x−8+4x= 7 =⇒7x−8 = 7
Step 4: Add 8 to both sides:
7x= 15
Step 5: Divide by 7:
x=15
7
Step 6: Substitute x=15
7back into y= 4 −2x:
y= 4 −215
7
Step 7: Simplify to find y:
y= 4 −30
7=28
7−30
7=−2
7
Step 8: Therefore, the solution to the system of equations is:
(x=15
7
y=−2
7
18
Question 21
Question
Solve the following system of equations:
(3x−2y= 1
2x+ 3y= 7
Solution
Step 1: Let’s solve the first equation for xin terms of y:
3x−2y= 1
3x= 2y+ 1
x=2y+ 1
3
Step 2: Substitute the expression for xinto the second equation:
22y+ 1
3+ 3y= 7
4y+ 2
3+ 3y= 7
4y+ 2 + 9y= 21
13y+ 2 = 21
13y= 19
y=19
13
Step 3: Substitute the value of yback into the expression for xto find its
value:
x=219
13 + 1
3
x=
38
13 + 1
3
x=
38+13
13
3
x=
51
13
3
x=51
39
x=17
13
Therefore, the solution to the system of equations is x=17
13 and y=19
13 .
19
Question 22
Question
Solve the following system of equations:
(2x+ 3y= 7
3x−2y= 4
Solution
Step 1: We will first solve the system of equations using the method of substi-
tution. We can rewrite the first equation as y=7−2x
3.
Step 2: Substitute y=7−2x
3into the second equation: 3x−27−2x
3= 4.
Step 3: Simplify the equation:
3x−14 −4x
3= 4
Step 4: Multiply through by 3 to clear the fraction:
9x−14 + 4x= 12
Step 5: Combine like terms:
13x−14 = 12
Step 6: Add 14 to both sides:
13x= 26
Step 7: Divide by 13:
x= 2
Step 8: Substitute x= 2 back into the equation 2x+ 3y= 7:
2(2) + 3y= 7
Step 9: Simplify:
4+3y= 7
Step 10: Subtract 4 from both sides:
3y= 3
Step 11: Divide by 3:
y= 1
Step 12: Therefore, the solution to the system of equations is x= 2 and
y= 1.
20
Question 23
Question
Solve the following system of equations:
(3x+ 2y= 5
2x−y= 4
Solution
Step 1: Solve the second equation for y:
2x−y= 4 =⇒ −y=−2x+ 4 =⇒y= 2x−4
Step 2: Substitute the expression for yinto the first equation and solve for
x:
3x+2(2x−4) = 5 =⇒3x+4x−8 = 5 =⇒7x−8 = 5 =⇒7x= 13 =⇒x=13
7
Step 3: Substitute the value of xback into the expression for yto solve for
y:
y= 2 13
7−4 = 26
7−4 = 26 −28
7=−2
7
Step 4: Therefore, the solution to the system of equations is (x, y) =
13
7,−2
7.
Question 24
Question
Solve the following system of equations:
(3x+ 4y= 7
2x−y= 5
Solution
Step 1: Solve the second equation for y.
2x−y= 5
−y= 5 −2x
y=−5+2x
21
Step 2: Substitute y=−5+2xinto the first equation.
3x+ 4(−5+2x)=7
3x−20 + 8x= 7
11x−20 = 7
11x= 27
x=27
11
Step 3: Substitute x=27
11 back into y=−5+2xto find y.
y=−5+227
11
y=−5 + 54
11
y=−55
11 +54
11
y=−1
11
Therefore, the solution to the system of equations is x=27
11 and y=−1
11 .
Question 25
Question
Solve the system of equations:
(3x−2y= 4
2x+ 3y= 1
Solution
Step 1: Multiply the first equation by 2 and the second equation by 3 to elim-
inate xwhen adding the equations. Step 2: Add the two resulting equations
to solve for y. Step 3: Substitute the value of yback into one of the original
equations to solve for x. Step 4: Verify the solution by substituting the values
of xand yinto both original equations.
22
Question 26
Question
Solve the following system of equations:
3x−4y+ 2z= 5
5x+ 2y−z=−1
x+ 3y−z= 4
Solution
Step 1: Multiply the second equation by 2 and add it to the first equation to
eliminate y:
3x−4y+ 2z= 5
12x+ 4y−2z=−2
x+ 3y−z= 4
Step 2: Add the equations together to solve for x:
16x= 7 ⇒x=7
16
Step 3: Substitute x=7
16 into the third equation to solve for y:
7
16 + 3y−z= 4 ⇒3y=57
16 +z
Step 4: Substitute x=7
16 and 3y=57
16 +zinto the first equation and solve
for z:
37
16−457
16 +z+ 2z= 5
⇒21
16 −57
4−4z+ 2z= 5
⇒ −57
4−2z=59
16 ⇒z=−227
32
Step 5: Finally, substitute x=7
16 and z=−227
32 into 3y=57
16 +zto solve
for y:
3y=57
16 −227
32
⇒3y=181
32 ⇒y=181
96
Therefore, the solution to the system of equations is:
x=7
16
y=181
96
z=−227
32
23
Question 27
Question
Solve the following system of equations:
(2x+ 3y= 7
3x−2y= 4
Solution
Step 1: We will first solve one of the equations for one variable and substitute
it into the other equation.
From 2x+ 3y= 7,
2x= 7 −3y,
x=7−3y
2.
Step 2: Substitute xinto the second equation.
37−3y
2−2y= 4,
21 −9y
2−2y= 4,
21 −9y−4y= 8,
21 −13y= 8,
−13y=−13,
y= 1.
Step 3: Substitute y= 1 into x=7−3y
2.
x=7−3(1)
2,
x=7−3
2,
x=4
2,
x= 2.
Therefore, the solution to the system of equations is x= 2 and y= 1.
24
Question 28
Question
Solve the following system of equations using the substitution method:
(3x−y= 4
x2+y2= 25
Solution
Step 1: Solve the first equation for yin terms of x.
3x−y= 4
y= 3x−4
Step 2: Substitute y= 3x−4 into the second equation.
x2+ (3x−4)2= 25
x2+ 9x2−24x+ 16 = 25
10x2−24x−9=0
Step 3: Solve the quadratic equation 10x2−24x−9 = 0 using the quadratic
formula.
x=−(−24) ±p(−24)2−4(10)(−9)
2(10)
x=24 ±√576 + 360
20
x=24 ±√936
20
x=24 ±30
20
Thus, x=54
20 =27
10 or x=−6
20 =−3
10 .
Step 4: Find the corresponding values of yusing y= 3x−4.
For x=27
10 :
y= 3 27
10−4
y=81
10 −40
10
y=41
10
So, one solution is x=27
10 and y=41
10 .
25
For x=−3
10 :
y= 3 −3
10−4
y=−9
10 −40
10
y=−49
10
So, another solution is x=−3
10 and y=−49
10 .
Therefore, the solutions to the system of equations are (x, y) = 27
10 ,41
10 and
(x, y) = −3
10 ,−49
10 .
Question 29
Question
Solve the following system of equations:
3x+ 2y−z= 6
x−y+z= 4
2x+ 3y−2z= 1
Solution
Step 1: Let’s rewrite the system of equations as an augmented matrix and row
reduce to solve for x,y, and z.
[ccc|c]3 2 −1 6
1−114
2 3 −2 1
Step 2: To simplify calculation, we will first multiply the second row by 3
and subtract the first row from it, and then multiply the third row by 2 and
subtract the first row from it.
[ccc|c]3 2 −1 6
0−5 4 12
0 5 −4−11
Step 3: Next, we will multiply the second row by −1/5 to simplify the
coefficient of y.
[ccc|c]3 2 −1 6
0 1 −0.8−2.4
0 5 −4−11
26
Step 4: Now we will add 5 times the second row to the third row to simplify
the coefficient of yin the third row.
[ccc|c]3 2 −1 6
0 1 −0.8−2.4
0 0 0 0
Step 5: We can see that the last row represents the equation 0 = 0, which
indicates that there are infinitely many solutions. Let’s express yin terms of z
and then substitute it back to find xand y. From the second row, we have:
y−0.8z=−2.4
y= 0.8z−2.4
Step 6: Substitute y= 0.8z−2.4 back into the first row to solve for x.
3x+ 2(0.8z−2.4) −z= 6
3x+ 1.6z−4.8−z= 6
3x+ 0.6z= 10.8
x= 3.6−0.2z
Therefore, the solutions to the system of equations are:
x= 3.6−0.2z, y = 0.8z−2.4, z =z
where zcan take any real value.
Question 30
Question
Solve the following system of equations:
2x−y+ 3z= 5
x+ 2y−z= 4
3x−2y+ 5z= 7
Solution
Step 1: Let’s use the method of substitution to solve this system of equations.
From the second equation, we can express xin terms of yand z:
x= 4 + z−2y
Step 2: Substitute the expression for xinto the first and third equations:
Substitute xin the first equation:
2(4 + z−2y)−y+ 3z= 5
27
8+2z−4y−y+ 3z= 5
5z−5y=−3
z−y=−3
5
Now substitute xin the third equation:
3(4 + z−2y)−2y+ 5z= 7
12 + 3z−6y−2y+ 5z= 7
8z−8y=−5
z−y=−5
8
Step 3: We now have a new system of equations to solve:
(z−y=−3
5
z−y=−5
8
Step 4: Subtract one equation from the other to eliminate y:
z−y−(z−y) = −3
5−(−5
8)
0 = −3
5+5
8
0 = −24
40 +25
40
0 = 1
40
Step 5: Since the equation 0 = 1
40 is not true, the system of equations is
inconsistent and has no solution.
Therefore, the system of equations has no solution.
Question 31
Question
Solve the following system of equations:
(2x+ 3y= 8
4x−y= 5
28
Solution
Step 1: Solve the second equation for y.
4x−y= 5
y= 4x−5
Step 2: Substitute yfrom the second equation into the first equation.
2x+ 3(4x−5) = 8
2x+ 12x−15 = 8
14x−15 = 8
14x= 23
x=23
14
Step 3: Substitute xinto the equation y= 4x−5 to find y.
y= 4 23
14−5
y=92
14 −5
y=92 −70
14
y=22
14
y=11
7
Therefore, the solution to the system of equations is x=23
14 and y=11
7.
Question 32
Question
Solve the following system of equations:
(3x+ 2y= 5
2x−3y= 1
Solution
Step 1: Let’s start by solving the first equation for x:
3x+ 2y= 5
3x= 5 −2y
x=5−2y
3
29
Step 2: Now let’s substitute the expression for xinto the second equation:
25−2y
3−3y= 1
10 −4y
3−3y= 1
10 −4y−9y= 3
10 −13y= 3
−13y=−7
y=−7
−13
y=7
13
Step 3: Next, substitute y=7
13 back into the expression for xfrom Step 1:
x=5−27
13
3
x=5−14
13
3
x=65 −14
39
x=51
39
x=17
13
Therefore, the solution to the system of equations is x=17
13 and y=7
13 .
Question 33
Question
Solve the following system of equations:
(3x−2y= 5
2x+ 3y= 4
Solution
Step 1: Let’s first multiply the first equation by 3 and the second equation by
2 to eliminate ywhen we add the equations.
Multiplying the first equation by 3 gives:
n9x−6y= 15
30
Multiplying the second equation by 2 gives:
n4x+ 6y= 8
Step 2: Now, add the modified equations to eliminate y:
9x−6y+ 4x+ 6y= 15 + 8
13x= 23
Step 3: Solve for x:
x=23
13
Step 4: Substitute xback into the first equation to solve for y:
3(23
13)−2y= 5
69
13 −2y= 5
−2y= 5 −69
13
−2y=65
13 −69
13
−2y=−4
13
y=2
13
Step 5: The solution to the system of equations is:
x=23
13, y =2
13
Question 34
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 8
31
Solution
Step 1: Let’s solve the second equation for yin terms of x.
(2x+ 3y= 7
4x−y= 8
Step 2: Rearrange the second equation to solve for y.
y= 4x−8
Step 3: Substitute y= 4x−8 into the first equation.
2x+ 3(4x−8) = 7
Step 4: Simplify the equation obtained.
2x+ 12x−24 = 7
14x−24 = 7
14x= 31
x=31
14
Step 5: Substitute x=31
14 back into y= 4x−8 to find the corresponding y
value.
y= 4 31
14−8
y=31
7−8
y=31
7−56
7
y=31 −56
7
y=−25
7
Step 6: Therefore, the solution to the system of equations is x=31
14 and y=−25
7.
Question 35
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 5
32
Question 2
Question
Solve the following system of equations:
(2x−3y= 7
4x+ 5y= 13
Solution
Step 1: Multiply the first equation by 2 to eliminate xwhen adding the equa-
tions. (4x−6y= 14
4x+ 5y= 13
Step 2: Subtract the first equation from the second to eliminate x.
11y=−1
Step 3: Solve for y.
y=−1
11
Step 4: Substitute yback into one of the original equations to solve for x.
2x−3(−1
11)=7
2x+3
11 = 7
2x=76
11
x=38
11
Step 5: The solution to the system is:
(x=38
11
y=−1
11
Question 3
Question
Solve the following system of equations:
(3x+ 2y= 10
x2+y2= 25
2
Solution
Step 1: We can start by isolating xin the first equation:
3x+ 2y= 10 =⇒3x= 10 −2y=⇒x=10 −2y
3
Step 2: Substitute the expression for xinto the second equation:
10 −2y
32
+y2= 25
Step 3: Simplifying the equation by expanding and combining like terms, we
get:
(10 −2y)2
9+y2= 25
Step 4: Further simplifying the equation by expanding and combining like
terms, we get:
100 −40y+ 4y2
9+y2= 25 =⇒100 −40y+ 4y2+ 9y2= 225
Step 5: Rearranging the terms, we have a quadratic equation:
13y2−40y−125 = 0
Step 6: Solve the quadratic equation for y. We can factor or use the quadratic
formula to find the solutions.
Step 7: Once we find the values for y, we can substitute them back into the
equation we found for x. This will give us the corresponding values for x.
Step 8: Finally, we check the solutions by substituting them into both orig-
inal equations to ensure they satisfy both equations.
Question 4
Question
Solve the following system of equations:
(2x+y= 5
3x−2y= 8
Solution
Step 1: We will solve the first equation for yin terms of x.
2x+y= 5
y= 5 −2x
3
Step 2: We will substitute the expression for yinto the second equation and
solve for x.3x−2(5 −2x)=8
3x−10 + 4x= 8
7x−10 = 8
7x= 18
x=18
7
Step 3: Now, we will substitute the value of xback into the first equation
to find y.
218
7+y= 5
36
7+y= 5
y= 5 −36
7
y=35
7−36
7
y=−1
7
Therefore, the solution to the system of equations is x=18
7and y=−1
7.
Question 5
Question
Solve the following system of equations:
(3x−2y= 5
2x+y= 3
Solution
Step 1: Start by solving the second equation for y.
2x+y= 3
y= 3 −2x
Step 2: Substitute this expression for yinto the first equation.
3x−2(3 −2x)=5
3x−6+4x= 5
7x= 11
x=11
7
4
Step 3: Substitute the value of xback into the equation y= 3 −2xto find
y.
y= 3 −211
7
y= 3 −22
7
y=1
7
So, the solution to the system of equations is x=11
7and y=1
7.
Question 6
Question
Solve the following system of equations:
2x−3y+z= 7
x+ 2y−2z=−1
3x−y+z= 5
Solution
Step 1: Write the system of equations in matrix form and represent it as an
augmented matrix.
2−3 1 7
1 2 −2−1
3−1 1 5
Step 2: Perform row operations to obtain row-echelon form.
1 2 −2−1
0−7 5 8
0−7 5 8
Step 3: Subtract the second row from the third row to get a new row.
1 2 −2−1
0−7 5 8
0 0 0 0
Step 4: Express the system of equations in its simplified form.
x+ 2y−2z=−1
−7y+ 5z= 8
0 = 0
5
Step 5: Solve the second equation for y.
−7y+ 5z= 8 =⇒y=5z−8
7
Step 6: Substitute the expression for yinto the first equation and solve for
x.
x+ 2 5z−8
7−2z=−1
Simplify the equation and express xin terms of z.
Step 7: Substitute the expressions for xand yback into one of the original
equations to solve for z.
Step 8: Once zis found, substitute back into the expressions for xand y
to obtain their values.
Question 7
Question
Solve the following system of equations:
3x+ 2y−z= 7
x−y+z=−2
2x+ 3y+z= 6
Solution
Step 1: Begin by writing the system of equations as an augmented matrix:
3 2 −1 7
1−1 1 −2
2 3 1 6
Step 2: Perform row operations to simplify the matrix.
3 2 −1 7
1−1 1 −2
2 3 1 6
→
1−1 1 −2
3 2 −1 7
2 3 1 6
Step 3: Use row operations to simplify the matrix further.
1−1 1 −2
3 2 −1 7
2 3 1 6
→
1−1 1 −2
0 5 −4 13
0 5 −1 10
Step 4: Continue to simplify the matrix by performing row operations.
1−1 1 −2
0 5 −4 13
0 5 −1 10
→
1−1 1 −2
0 5 −4 13
0 0 3 −3
6
Step 5: Use the row echelon form of the matrix to back-substitute and find
the values of the variables. From the last row, we find that 3z=−3⇒z=−1.
Substituting z=−1 into the middle row equation, we get 5y+ 4 = 13 ⇒y= 3.
Finally, substituting y= 3 and z=−1 into the top row equation, we get
x−3−1 = −2⇒x= 0.
Therefore, the solution to the system of equations is x= 0, y= 3, z=−1.
Question 8
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−5y=−3
Solution
Step 1: Solve the first equation for xin terms of y.
2x+ 3y= 7 ⇒2x= 7 −3y⇒x=7−3y
2
Step 2: Substitute xin the second equation.
47−3y
2−5y=−3
Step 3: Simplify and solve for y.
14 −6y−5y=−3⇒14 −11y=−3⇒ −11y=−17 ⇒y=17
11
Step 4: Substitute yback to find x.
x=7−3(17
11 )
2
Step 5: Calculate the value of x.
x=7−51
11
2=77 −51
22 =26
22 =13
11
Step 6: The solution to the system of equations is x=13
11 and y=17
11 .
Question 9
Question
Solve the following system of equations:
(2x+ 3y= 8
3x−4y= 5
7
Solution
Step 1: Let’s start by multiplying the first equation by 3 and the second equation
by 2 to eliminate xwhen we add the equations together.
(6x+ 9y= 24
6x−8y= 10
Step 2: Now, let’s subtract the second equation from the first equation to
eliminate x.
17y= 14
Step 3: Solve for y.
y=14
17
Step 4: Substitute y=14
17 back into one of the original equations to solve
for x. Let’s use the first equation.
2x+ 3 14
17= 8
Step 5: Simplify the equation and solve for x.
2x+42
17 = 8
Step 6: Solve for x.
x=70
17
Therefore, the solution to the system of equations is x=70
17 and y=14
17 .
Question 10
Question
Solve the following system of equations:
2x−3y+z= 7
3x+ 2y−2z=−8
x+y+z= 1
Solution
Step 1: We will use the method of substitution to solve this system. Let’s start
by isolating one variable in the third equation. Subtracting yand zfrom both
sides of the third equation, we get:
x= 1 −y−z
8
Step 2: Now substitute x= 1 −y−zinto the first and second equations to
create two new equations with yand zas variables. Substitute xinto the first
equation:
2(1 −y−z)−3y+z= 7
2−2y−2z−3y+z= 7
2−5y−z= 7
5y+z=−5
This is our new equation 1.
Step 3: Substitute x= 1 −y−zinto the second equation:
3(1 −y−z)+2y−2z=−8
3−3y−3z+ 2y−2z=−8
3−y−5z=−8
y+ 5z= 11
This is our new equation 2.
Step 4: Now we have a system of two equations in two variables yand z:
(5y+z=−5
y+ 5z= 11
Step 5: We can now solve this new system of equations by multiplying the
first equation by 5 and adding the equations to eliminate y:
25y+ 5z=−25
5y+z=−5
Adding the two equations gives:
30y=−30
y=−1
Step 6: Finally, substitute y=−1 back into equation 1:
5(−1) + z=−5
−5 + z=−5
z= 0
Step 7: Now that we have y=−1 and z= 0, substitute these values into
x= 1 −y−zto find x:
x= 1 −(−1) −0
9
x= 2
Step 8: Therefore, the solution to the system of equations is:
x= 2
y=−1
z= 0
Question 11
Question
Solve the following system of equations:
(2x+ 3y= 7
4x+ 5y= 11
Solution
Step 1: Let’s start by multiplying the first equation by 2 to match the coeffi-
cients of x in both equations:
(4x+ 6y= 14
4x+ 5y= 11
Step 2: Now, subtract the second equation from the first equation to elim-
inate x: (4x+ 6y)−(4x+ 5y) = 14 −11
6y−5y= 3
y= 3
Step 3: Substitute the value of y back into one of the original equations to
solve for x. Let’s use the first equation:
2x+ 3(3) = 7
Step 4: Solve for x:
2x+ 9 = 7
2x=−2
x=−1
Step 5: Therefore, the solution to the system of equations is x=−1 and
y= 3.
10
Question 12
Question
Solve the following system of equations:
(3x+ 2y= 7
4x−y= 5
Solution
Step 1: We will solve the second equation for y:
y= 4x−5
Step 2: Substitute the expression for yinto the first equation:
3x+ 2(4x−5) = 7
Step 3: Simplify the equation:
3x+ 8x−10 = 7
11x−10 = 7
Step 4: Solve for x:
11x= 17
x=17
11
Step 5: Substitute the value of xback into the equation for y:
y= 4 17
11−5
y=68
11 −55
11
y=13
11
Step 6: Therefore, the solution to the system of equations is:
x=17
11, y =13
11
Question 13
Question
Solve the following system of equations:
3x+ 2y+z= 9
4x−y+ 2z= 7
x+ 3y−z= 4
11
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 z:
3x+ 2y+z= 9
z= 9 −3x−2y
Step 2: Now, substitute z= 9 −3x−2yinto the second equation:
4x−y+ 2(9 −3x−2y)=7
4x−y+ 18 −6x−4y= 7
−2x−5y=−11
2x+ 5y= 11 (⋆)
Step 3: Next, substitute z= 9 −3x−2yinto the third equation:
x+ 3y−(9 −3x−2y)=4
x+ 3y−9+3x+ 2y= 4
4x+ 5y= 13 (⋆⋆)
Step 4: Now, we have the system of equations:
(2x+ 5y= 11
4x+ 5y= 13
Step 5: Subtract the first equation from the second equation to get:
(4x+ 5y)−(2x+ 5y) = 13 −11
2x= 2
x= 1
Step 6: Substitute x= 1 into equation (⋆):
2(1) + 5y= 11
2+5y= 11
5y= 9
y=9
5
Step 7: Substitute x= 1 and y=9
5into one of the original equations (e.g.
the first equation) to solve for z:
3(1) + 2 9
5+z= 9
3 + 18
5+z= 9
z= 9 −3−18
5
z=12
5
12
Therefore, the solution to the system of equations is x= 1, y =9
5, z =12
5.
Question 14
Question
Solve the following system of equations:
(2x+ 3y= 5
4x−5y= 7
Solution
Step 1: Multiply the first equation by 2 to eliminate xwhen adding the two
equations. This gives us
(4x+ 6y= 10
4x−5y= 7
Step 2: Subtract the second equation from the first equation to eliminate x:
(4x+ 6y)−(4x−5y) = 10 −7
11y= 3
y=3
11
Step 3: Substitute the value of yback into the first equation to solve for x:
2x+ 3 3
11= 5
2x+9
11 = 5
2x= 5 −9
11
2x=55
11 −9
11
2x=46
11
x=46
22
x=23
11
Therefore, the solution to the system of equations is x=23
11 and y=3
11 .
13
Question 15
Question
Solve the following system of equations:
(2x+ 3y= 11
3x−2y= 4
Solution
Step 1: Start by multiplying the first equation by 3 and the second equation by
2 to get the coefficients of xto be the same:
(6x+ 9y= 33
6x−4y= 8
Step 2: Now, subtract the second equation from the first equation:
13y= 25
Step 3: Solve for y:
y=25
13
Step 4: Substitute yback into the first equation to solve for x:
2x+ 3 25
13= 11
Step 5: Simplify the equation and solve for x:
2x+75
13 = 11
Step 6:
2x= 11 −75
13 =143
13 −75
13 =68
13
Step 7:
x=68
13∇ · 2 = 68
13 ×1
2=34
13
So, the solution to the system of equations is x=34
13 and y=25
13 .
Question 16
Question
Solve the following system of equations:
(5x+ 3y= 16
3x−2y= 4
14
Solution
Step 1: Multiply the first equation by 2 and the second equation by 3 to create
opposite coefficients for yand eliminate ywhen combining the equations.
(10x+ 6y= 32
9x−6y= 12
Step 2: Add the new equations together to eliminate y.
(10x+ 6y) + (9x−6y) = 32 + 12
19x= 44
x=44
19
Step 3: Substitute xback into one of the original equations to solve for y.
Using the first equation:
544
19+ 3y= 16
220
19 + 3y= 16
3y= 16 −220
19
3y=224
19
y=224
57
Therefore, the solution to the system of equations is x=44
19 and y=224
57 .
Question 17
Question
Solve the following system of equations:
(2x+ 3y= 5
3x−y= 7
Solution
Step 1: Solve the second equation for yin terms of x.
3x−y= 7
y= 3x−7
15
Step 2: Substitute the expression for yinto the first equation.
2x+ 3(3x−7) = 5
2x+ 9x−21 = 5
11x= 26
x=26
11 =2
1
Step 3: Substitute the value of xback into the second equation to solve for
y.
3(2
1)−y= 7
6−y= 7
−y= 1
y=−1
Therefore, the solution to the system of equations is x=2
1and y=−1.
Question 18
Question
Solve the following system of equations:
(2x−3y= 5
x2+y2= 25
Solution
Step 1: Multiply the first equation by 2 to simplify the system:
(4x−6y= 10
x2+y2= 25
Step 2: Square the first equation to eliminate y:
((4x−6y)2= 100
x2+y2= 25
Step 3: Expand and simplify the squared equation:
(16x2−48xy + 36y2= 100
x2+y2= 25
16
Step 4: Substitute y2= 25 −x2from the second equation into the first
equation:
(16x2−48x(5 −x) + 36(25 −x2) = 100
x2+ 25 −x2= 25
Step 5: Simplify the system of equations:
(16x2−240x+ 900 −36x2= 100
25 = 25
Step 6: Combine like terms and solve for x:
(−20x+ 800 = 100
25 = 25
Step 7: Solve for x:
(−20x=−700
25 = 25
Step 8: Find the value of x:
x=−700
−20 = 35
Step 9: Substitute x= 35 back into x2+y2= 25 to solve for y:
352+y2= 25
1225 + y2= 25
y2=−1200
Step 10: There are no real solutions for ysince y2cannot be negative.
Therefore, the system has no real solutions.
Therefore, the system of equations has no real solutions.
Question 19
Question
Solve the following system of equations:
2x−3y+ 5z= 4
x+y−z= 3
4x−5y+z= 7
Solution
Step-by-step solution goes here.
17
Question 20
Question
Solve the following system of equations:
(3x−2y= 7
2x+y= 4
Solution
To solve this system of equations, we can use the method of substitution or
elimination.
Step 1: Let’s use the method of substitution. Solve the second equation for
yin terms of x:
2x+y= 4 =⇒y= 4 −2x
Step 2: Substitute y= 4 −2xinto the first equation:
3x−2(4 −2x) = 7
Step 3: Expand and simplify the equation:
3x−8+4x= 7 =⇒7x−8 = 7
Step 4: Add 8 to both sides:
7x= 15
Step 5: Divide by 7:
x=15
7
Step 6: Substitute x=15
7back into y= 4 −2x:
y= 4 −215
7
Step 7: Simplify to find y:
y= 4 −30
7=28
7−30
7=−2
7
Step 8: Therefore, the solution to the system of equations is:
(x=15
7
y=−2
7
18
Question 21
Question
Solve the following system of equations:
(3x−2y= 1
2x+ 3y= 7
Solution
Step 1: Let’s solve the first equation for xin terms of y:
3x−2y= 1
3x= 2y+ 1
x=2y+ 1
3
Step 2: Substitute the expression for xinto the second equation:
22y+ 1
3+ 3y= 7
4y+ 2
3+ 3y= 7
4y+ 2 + 9y= 21
13y+ 2 = 21
13y= 19
y=19
13
Step 3: Substitute the value of yback into the expression for xto find its
value:
x=219
13 + 1
3
x=
38
13 + 1
3
x=
38+13
13
3
x=
51
13
3
x=51
39
x=17
13
Therefore, the solution to the system of equations is x=17
13 and y=19
13 .
19
Question 22
Question
Solve the following system of equations:
(2x+ 3y= 7
3x−2y= 4
Solution
Step 1: We will first solve the system of equations using the method of substi-
tution. We can rewrite the first equation as y=7−2x
3.
Step 2: Substitute y=7−2x
3into the second equation: 3x−27−2x
3= 4.
Step 3: Simplify the equation:
3x−14 −4x
3= 4
Step 4: Multiply through by 3 to clear the fraction:
9x−14 + 4x= 12
Step 5: Combine like terms:
13x−14 = 12
Step 6: Add 14 to both sides:
13x= 26
Step 7: Divide by 13:
x= 2
Step 8: Substitute x= 2 back into the equation 2x+ 3y= 7:
2(2) + 3y= 7
Step 9: Simplify:
4+3y= 7
Step 10: Subtract 4 from both sides:
3y= 3
Step 11: Divide by 3:
y= 1
Step 12: Therefore, the solution to the system of equations is x= 2 and
y= 1.
20
Question 23
Question
Solve the following system of equations:
(3x+ 2y= 5
2x−y= 4
Solution
Step 1: Solve the second equation for y:
2x−y= 4 =⇒ −y=−2x+ 4 =⇒y= 2x−4
Step 2: Substitute the expression for yinto the first equation and solve for
x:
3x+2(2x−4) = 5 =⇒3x+4x−8 = 5 =⇒7x−8 = 5 =⇒7x= 13 =⇒x=13
7
Step 3: Substitute the value of xback into the expression for yto solve for
y:
y= 2 13
7−4 = 26
7−4 = 26 −28
7=−2
7
Step 4: Therefore, the solution to the system of equations is (x, y) =
13
7,−2
7.
Question 24
Question
Solve the following system of equations:
(3x+ 4y= 7
2x−y= 5
Solution
Step 1: Solve the second equation for y.
2x−y= 5
−y= 5 −2x
y=−5+2x
21
Step 2: Substitute y=−5+2xinto the first equation.
3x+ 4(−5+2x)=7
3x−20 + 8x= 7
11x−20 = 7
11x= 27
x=27
11
Step 3: Substitute x=27
11 back into y=−5+2xto find y.
y=−5+227
11
y=−5 + 54
11
y=−55
11 +54
11
y=−1
11
Therefore, the solution to the system of equations is x=27
11 and y=−1
11 .
Question 25
Question
Solve the system of equations:
(3x−2y= 4
2x+ 3y= 1
Solution
Step 1: Multiply the first equation by 2 and the second equation by 3 to elim-
inate xwhen adding the equations. Step 2: Add the two resulting equations
to solve for y. Step 3: Substitute the value of yback into one of the original
equations to solve for x. Step 4: Verify the solution by substituting the values
of xand yinto both original equations.
22
Question 26
Question
Solve the following system of equations:
3x−4y+ 2z= 5
5x+ 2y−z=−1
x+ 3y−z= 4
Solution
Step 1: Multiply the second equation by 2 and add it to the first equation to
eliminate y:
3x−4y+ 2z= 5
12x+ 4y−2z=−2
x+ 3y−z= 4
Step 2: Add the equations together to solve for x:
16x= 7 ⇒x=7
16
Step 3: Substitute x=7
16 into the third equation to solve for y:
7
16 + 3y−z= 4 ⇒3y=57
16 +z
Step 4: Substitute x=7
16 and 3y=57
16 +zinto the first equation and solve
for z:
37
16−457
16 +z+ 2z= 5
⇒21
16 −57
4−4z+ 2z= 5
⇒ −57
4−2z=59
16 ⇒z=−227
32
Step 5: Finally, substitute x=7
16 and z=−227
32 into 3y=57
16 +zto solve
for y:
3y=57
16 −227
32
⇒3y=181
32 ⇒y=181
96
Therefore, the solution to the system of equations is:
x=7
16
y=181
96
z=−227
32
23
Question 27
Question
Solve the following system of equations:
(2x+ 3y= 7
3x−2y= 4
Solution
Step 1: We will first solve one of the equations for one variable and substitute
it into the other equation.
From 2x+ 3y= 7,
2x= 7 −3y,
x=7−3y
2.
Step 2: Substitute xinto the second equation.
37−3y
2−2y= 4,
21 −9y
2−2y= 4,
21 −9y−4y= 8,
21 −13y= 8,
−13y=−13,
y= 1.
Step 3: Substitute y= 1 into x=7−3y
2.
x=7−3(1)
2,
x=7−3
2,
x=4
2,
x= 2.
Therefore, the solution to the system of equations is x= 2 and y= 1.
24
Question 28
Question
Solve the following system of equations using the substitution method:
(3x−y= 4
x2+y2= 25
Solution
Step 1: Solve the first equation for yin terms of x.
3x−y= 4
y= 3x−4
Step 2: Substitute y= 3x−4 into the second equation.
x2+ (3x−4)2= 25
x2+ 9x2−24x+ 16 = 25
10x2−24x−9=0
Step 3: Solve the quadratic equation 10x2−24x−9 = 0 using the quadratic
formula.
x=−(−24) ±p(−24)2−4(10)(−9)
2(10)
x=24 ±√576 + 360
20
x=24 ±√936
20
x=24 ±30
20
Thus, x=54
20 =27
10 or x=−6
20 =−3
10 .
Step 4: Find the corresponding values of yusing y= 3x−4.
For x=27
10 :
y= 3 27
10−4
y=81
10 −40
10
y=41
10
So, one solution is x=27
10 and y=41
10 .
25
For x=−3
10 :
y= 3 −3
10−4
y=−9
10 −40
10
y=−49
10
So, another solution is x=−3
10 and y=−49
10 .
Therefore, the solutions to the system of equations are (x, y) = 27
10 ,41
10 and
(x, y) = −3
10 ,−49
10 .
Question 29
Question
Solve the following system of equations:
3x+ 2y−z= 6
x−y+z= 4
2x+ 3y−2z= 1
Solution
Step 1: Let’s rewrite the system of equations as an augmented matrix and row
reduce to solve for x,y, and z.
[ccc|c]3 2 −1 6
1−114
2 3 −2 1
Step 2: To simplify calculation, we will first multiply the second row by 3
and subtract the first row from it, and then multiply the third row by 2 and
subtract the first row from it.
[ccc|c]3 2 −1 6
0−5 4 12
0 5 −4−11
Step 3: Next, we will multiply the second row by −1/5 to simplify the
coefficient of y.
[ccc|c]3 2 −1 6
0 1 −0.8−2.4
0 5 −4−11
26
Step 4: Now we will add 5 times the second row to the third row to simplify
the coefficient of yin the third row.
[ccc|c]3 2 −1 6
0 1 −0.8−2.4
0 0 0 0
Step 5: We can see that the last row represents the equation 0 = 0, which
indicates that there are infinitely many solutions. Let’s express yin terms of z
and then substitute it back to find xand y. From the second row, we have:
y−0.8z=−2.4
y= 0.8z−2.4
Step 6: Substitute y= 0.8z−2.4 back into the first row to solve for x.
3x+ 2(0.8z−2.4) −z= 6
3x+ 1.6z−4.8−z= 6
3x+ 0.6z= 10.8
x= 3.6−0.2z
Therefore, the solutions to the system of equations are:
x= 3.6−0.2z, y = 0.8z−2.4, z =z
where zcan take any real value.
Question 30
Question
Solve the following system of equations:
2x−y+ 3z= 5
x+ 2y−z= 4
3x−2y+ 5z= 7
Solution
Step 1: Let’s use the method of substitution to solve this system of equations.
From the second equation, we can express xin terms of yand z:
x= 4 + z−2y
Step 2: Substitute the expression for xinto the first and third equations:
Substitute xin the first equation:
2(4 + z−2y)−y+ 3z= 5
27
8+2z−4y−y+ 3z= 5
5z−5y=−3
z−y=−3
5
Now substitute xin the third equation:
3(4 + z−2y)−2y+ 5z= 7
12 + 3z−6y−2y+ 5z= 7
8z−8y=−5
z−y=−5
8
Step 3: We now have a new system of equations to solve:
(z−y=−3
5
z−y=−5
8
Step 4: Subtract one equation from the other to eliminate y:
z−y−(z−y) = −3
5−(−5
8)
0 = −3
5+5
8
0 = −24
40 +25
40
0 = 1
40
Step 5: Since the equation 0 = 1
40 is not true, the system of equations is
inconsistent and has no solution.
Therefore, the system of equations has no solution.
Question 31
Question
Solve the following system of equations:
(2x+ 3y= 8
4x−y= 5
28
Solution
Step 1: Solve the second equation for y.
4x−y= 5
y= 4x−5
Step 2: Substitute yfrom the second equation into the first equation.
2x+ 3(4x−5) = 8
2x+ 12x−15 = 8
14x−15 = 8
14x= 23
x=23
14
Step 3: Substitute xinto the equation y= 4x−5 to find y.
y= 4 23
14−5
y=92
14 −5
y=92 −70
14
y=22
14
y=11
7
Therefore, the solution to the system of equations is x=23
14 and y=11
7.
Question 32
Question
Solve the following system of equations:
(3x+ 2y= 5
2x−3y= 1
Solution
Step 1: Let’s start by solving the first equation for x:
3x+ 2y= 5
3x= 5 −2y
x=5−2y
3
29
Step 2: Now let’s substitute the expression for xinto the second equation:
25−2y
3−3y= 1
10 −4y
3−3y= 1
10 −4y−9y= 3
10 −13y= 3
−13y=−7
y=−7
−13
y=7
13
Step 3: Next, substitute y=7
13 back into the expression for xfrom Step 1:
x=5−27
13
3
x=5−14
13
3
x=65 −14
39
x=51
39
x=17
13
Therefore, the solution to the system of equations is x=17
13 and y=7
13 .
Question 33
Question
Solve the following system of equations:
(3x−2y= 5
2x+ 3y= 4
Solution
Step 1: Let’s first multiply the first equation by 3 and the second equation by
2 to eliminate ywhen we add the equations.
Multiplying the first equation by 3 gives:
n9x−6y= 15
30
Multiplying the second equation by 2 gives:
n4x+ 6y= 8
Step 2: Now, add the modified equations to eliminate y:
9x−6y+ 4x+ 6y= 15 + 8
13x= 23
Step 3: Solve for x:
x=23
13
Step 4: Substitute xback into the first equation to solve for y:
3(23
13)−2y= 5
69
13 −2y= 5
−2y= 5 −69
13
−2y=65
13 −69
13
−2y=−4
13
y=2
13
Step 5: The solution to the system of equations is:
x=23
13, y =2
13
Question 34
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 8
31
Solution
Step 1: Let’s solve the second equation for yin terms of x.
(2x+ 3y= 7
4x−y= 8
Step 2: Rearrange the second equation to solve for y.
y= 4x−8
Step 3: Substitute y= 4x−8 into the first equation.
2x+ 3(4x−8) = 7
Step 4: Simplify the equation obtained.
2x+ 12x−24 = 7
14x−24 = 7
14x= 31
x=31
14
Step 5: Substitute x=31
14 back into y= 4x−8 to find the corresponding y
value.
y= 4 31
14−8
y=31
7−8
y=31
7−56
7
y=31 −56
7
y=−25
7
Step 6: Therefore, the solution to the system of equations is x=31
14 and y=−25
7.
Question 35
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 5
32
Solution
Step 1: We will first solve the second equation for yin terms of x.
4x−y= 5
y= 4x−5
Step 2: Next, we will substitute yin the first equation with 4x−5 to form
one equation with only x.
2x+ 3(4x−5) = 7
2x+ 12x−15 = 7
14x= 22
x=22
14
x=11
7
Step 3: Now that we have found the value of x, we can substitute it back
into the equation y= 4x−5 to find the corresponding value of 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.
33
