MATH 114 - QUANTITATIVE
REASONING - Systems of Equations
Question Bank - Set 1
Liberty University
Question 1
Question
Solve the following system of equations:
(2x−3y= 1
4x+ 5y= 7
Solution
Step 1: Multiply the first equation by 4 and the second equation by 2 to make
the coefficients of xthe same:
(8x−12y= 4
8x+ 10y= 14
Step 2: Subtract the first equation from the second equation to eliminate x:
(8x+ 10y)−(8x−12y) = 14 −4
Simplify to get:
22y= 10
Step 3: Solve for y:
y=10
22 =5
11
Step 4: Substitute the value of yback into one of the original equations, for
example, the first equation, to solve for x:
2x−3( 5
11)=1
Solve for x:
2x−15
11 = 1
2x= 1 + 15
11 =11
11 +15
11 =26
11
x=13
11
Step 5: Therefore, the solution to the system of equations is:
(x=13
11
y=5
11
Question 2
Question
Solve the following system of equations:
(3x−y= 5
x2+y2= 25
Solution
Step 1: Solve the first equation for yin terms of x.
3x−y= 5 =⇒y= 3x−5
Step 2: Substitute y= 3x−5 into the second equation.
x2+ (3x−5)2= 25
Step 3: Expand and simplify the equation.
x2+ 9x2−30x+ 25 = 25
10x2−30x= 0
10x(x−3) = 0
Step 4: Solve for xby setting each factor to zero.
10x= 0 =⇒x= 0
x−3 = 0 =⇒x= 3
Step 5: Now, substitute the values of xback into the equation y= 3x−5
to find the corresponding values of y. When x= 0:
y= 3(0) −5 = −5 =⇒(x, y) = (0,−5)
When x= 3:
y= 3(3) −5 = 4 =⇒(x, y) = (3,4)
Step 6: The solution to the system of equations is (0,−5) and (3,4).
2
Question 3
Question
Solve the following system of equations:
(3x+ 5y= 17
2x−4y= 6
Solution
Step 1: We can solve this system of equations using the method of elimination.
Let’s start by multiplying the first equation by 2 and the second equation by 3
to create oppositely signed coefficients for x.
Step 2:
(6x+ 10y= 34
6x−12y= 18
Step 3: Now, we subtract the second equation from the first equation to
eliminate x:(6x+ 10y)−(6x−12y) = 34 −18
22y= 16
y=16
22
y=8
11
Step 4: Next, we substitute the value of yback into the first equation to
solve for x:
3x+ 5( 8
11) = 17
3x+40
11 = 17
3x= 17 −40
11
3x=187
11
x=187
33
Step 5: Therefore, the solution to the system of equations is:
(x=187
33
y=8
11
3
Question 4
Question
Solve the following system of equations:
(3x+ 2y= 7
4x+ 3y= 9
Solution
Step 1: We can solve the system of equations using the method of substitution.
Let’s solve the first equation for x: From the first equation 3x+ 2y= 7, we have
3x= 7 −2y
x=7−2y
3
Step 2: Now, substitute xin terms of yinto the second equation. Substitute
x=7−2y
3into the second equation 4x+ 3y= 9:
47−2y
3+ 3y= 9
Step 3: Simplify the equation:
28 −8y
3+ 3y= 9
28 −8y+ 9y= 27
y=−1
Step 4: Substitute y=−1 back into x=7−2y
3to find x:
x=7−2(−1)
3
x=9
3
x= 3
Step 5: Therefore, the solution to the system of equations is x= 3 and
y=−1.
Question 5
Question
Solve the following system of equations:
(2x−3y= 5
4x+y= 1
4
Solution
Step 1: Start by solving the second equation for yin terms of x.
4x+y= 1
y=−4x+ 1
Step 2: Substitute y=−4x+ 1 into the first equation to solve for x.
2x−3(−4x+ 1) = 5
2x+ 12x−3=5
14x−3=5
14x= 8
x=8
14
x=4
7
Step 3: Substitute x=4
7back into y=−4x+ 1 to find y.
y=−44
7+ 1
y=−16
7+ 1
y=−16
7+7
7
y=−9
7
Therefore, the solution to the system of equations is (x, y) = 4
7,−9
7.
Question 6
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 5
Solution
Step 1: Let’s solve the second equation for y.
4x−y= 5
−y= 5 −4x
y=−5+4x
5
Step 2: Substitute yfrom the second equation into the first equation.
2x+ 3(−5+4x)=7
2x−15 + 12x= 7
14x−15 = 7
14x= 22
x=22
14
x=11
7
Step 3: Now, substitute xback into the second equation to solve for y.
411
7−y= 5
44
7−y= 5
−y= 5 −44
7
−y=35 −44
7
−y=−9
7
y=9
7
Step 4: Therefore, the solution to the system of equations is x=11
7and
y=9
7.
Question 7
Question
Solve the following system of equations:
2x−3y+z= 7
x+ 2y−2z= 1
3x−5y−3z= 4
Solution
Step 1: We will begin by writing the system of equations in matrix form. Let
A=
2−3 1
1 2 −2
3−5−3
6
and
X=
x
y
z
,
where Xis the column matrix containing the variables x,y, and z, and rewrite
the system as AX=B, where B=
7
1
4
.
Step 2: Next, we will find the inverse of matrix Aby row reducing the
augmented matrix [A|I]:
2−3 1 |100
1 2 −2|010
3−5−3|001
Row reducing gives us
1 0 −5| −3 2 6
0 1 −4|2−1 3
0 0 0 |13 −5−9
Since the augmented matrix is not in reduced row-echelon form, the system
of equations is inconsistent, meaning there is no solution.
Question 8
Question
Solve the following system of equations:
2x−y+ 3z= 4
x+ 2y−z= 1
3x−2y+ 2z= 7
Solution
Step 1: Let’s first simplify the system by writing it as an augmented matrix and
then perform row operations to put it in row-echelon form.
2−1 3 |4
1 2 −1|1
3−2 2 |7
Step 2: We’ll start by subtracting 0.5×row 1 from row 2, and 1.5×row 1
from row 3.
2−1 3 |4
0 2.5−2.5| −1
0−3.5−2.5|1
7
Step 3: Next, we’ll multiply row 2 by 2
5to simplify calculations, and then
add 1.4×row 2 to row 3.
2−1 3 |4
0 1 −1| −0.4
0 0 −6|0.6
Step 4: We’ll divide row 3 by −6 to get the reduced row-echelon form.
2−1 3 |4
0 1 −1| −0.4
0 0 1 | −0.1
Step 5: Now, we will back-substitute to find the values of x,y, and z. From
the third row, z=−0.1. Substitute zinto the second row: y−1 = −0.4⇒y=
0.6. Substitute yand zinto the first row: 2x+ 0.6 + 0.3=4⇒2x= 3.1⇒x=
1.55.
Therefore, the solution to the system of equations is x= 1.55, y= 0.6, and
z=−0.1.
Question 9
Question
Find the solution to the following system of equations:
(2x+ 3y= 11
3x−2y= 1
Solution
Step 1: Let’s solve the first equation for xin terms of y.
2x+ 3y= 11
2x= 11 −3y
x=11 −3y
2
8
Step 2: Substitute xin the second equation.
311 −3y
2−2y= 1
33 −9y
2−2y= 1
33 −9y−4y= 2
33 −13y= 2
−13y=−31
y=31
13
y= 3
Step 3: Substitute y= 3 back into the first equation to find x.
2x+ 3(3) = 11
2x+ 9 = 11
2x= 2
x= 1
Thus, the solution to the system of equations is x= 1 and y= 3.
Question 10
Question
Solve the following system of equations:
(3x+ 2y= 7
2x−y= 1
Solution
Step 1: Solve the second equation for y:
2x−y= 1
y= 2x−1
Step 2: Substitute y= 2x−1 into the first equation and solve for x:
3x+ 2(2x−1) = 7
3x+ 4x−2=7
7x= 9
x=9
7
9
Step 3: Substitute x=9
7back into y= 2x−1 to solve for y:
y= 2 9
7−1
y=18
7−1
y=18
7−7
7
y=11
7
Therefore, the solution to the system of equations is x=9
7and y=11
7.
Question 11
Question
Solve the system of equations:
(2x−3y=−1
3x+ 4y= 9
Solution
Step 1: Multiply both sides of the first equation by 3 and the second equation
by 2 in order to eliminate ywhen adding the two equations together.
(6x−9y=−3
6x+ 8y= 18
Step 2: Subtract the first equation from the second equation to eliminate
x.
17y= 21
Step 3: Solve for yby dividing both sides by 17.
y=21
17
Step 4: Substitute the value of yback into one of the original equations to
solve for x. Using the first equation:
2x−321
17=−1
Step 5: Solve for x.
2x−63
17 =−1
10
2x=−1 + 63
17
2x=−17 + 63
17
2x=−80
17
x=−40
17
Step 6: The solution to the system of equations is x=−40
17 and y=21
17 .
Question 12
Question
Solve the following system of equations:
(3x+ 2y= 5
x2+y2= 10
Solution
Step 1: Let’s start by solving the first equation for yin terms of x.
3x+ 2y= 5
2y= 5 −3x
y=5−3x
2
Step 2: Substitute yby 5−3x
2in the second equation.
x2+5−3x
22
= 10
x2+(5 −3x)2
4= 10
x2+25 −30x+ 9x2
4= 10
4x2+ 25 −30x+ 9x2= 40
13x2−30x−15 = 0
11
Step 3: Solve the quadratic equation 13x2−30x−15 = 0.
x=−(−30) ±p(−30)2−4(13)(−15)
2(13)
x=30 ±√900 + 780
26
x=30 ±√1680
26
x=30 ±4√105
26
x=15 ±2√105
13
Step 4: Substitute the values of xback into the equation y=5−3x
2to find
the corresponding yvalues. Therefore, the solutions to the system of equations
are: 15 + 2√105
13 ,5−3(15+2√105
13 )
2!
and 15 −2√105
13 ,5−3(15−2√105
13 )
2!
Question 13
Question
Consider the following system of equations:
(2x−3y= 5
4x+ky = 11
Find the value(s) of kthat will make this system of equations have a unique
solution.
Solution
Step 1: To find the value(s) of kthat will make this system have a unique
solution, we need to ensure that the two equations represent distinct lines. This
means the lines must not be parallel or coincide.
Step 2: Let’s first write the second equation in slope-intercept form by solv-
ing for y:
4x+ky = 11 =⇒ky =−4x+ 11 =⇒y=−4
kx+11
k
12
Step 3: For the system of equations to have a unique solution, the slopes of
the two lines formed by the equations must not be equal. Therefore, the slope
of the first line 2x−3y= 5 and the slope of the second line −4
kx+11
kshould
not be equal.
Step 4: The slope of the first line 2x−3y= 5 is 2
3. So, for a unique solution
to exist, the slope of the second line −4
kshould not be equal to 2
3. Thus, −4
k=2
3.
Step 5: Solving −4
k=2
3for k, we get:
−4
k=2
3=⇒k=−6
Step 6: Therefore, for the system of equations to have a unique solution, k
must not be equal to −6. Thus, the value of kthat will make this system have
a unique solution is k=−6.
Question 14
Question
Solve the following system of equations:
(4x+ 3y= 7
2x−y= 3
Solution
Step 1: Solve the second equation for y:
2x−y= 3
y= 2x−3
Step 2: Substitute y= 2x−3 into the first equation and solve for x:
4x+ 3(2x−3) = 7
4x+ 6x−9=7
10x−9=7
10x= 16
x=16
10
x= 1.6
Step 3: Substitute x= 1.6 into y= 2x−3 to find y:
y= 2(1.6) −3
y= 3.2−3
y= 0.2
Therefore, the solution to the system of equations is x= 1.6 and y= 0.2.
13
Question 15
Question
Solve the following system of equations for xand y:
(3x+ 4y= 5
5x−y= 7
Solution
Step 1: Multiply the second equation by 4 to eliminate y:
(3x+ 4y= 5
20x−4y= 28
Step 2: Add the equations together to eliminate y:
23x= 33 ⇒x=33
23
Step 3: Substitute xback into the first equation to solve for y:
333
23+ 4y= 5 ⇒y=−7
23
Step 4: The solution to the system of equations is x=33
23 and y=−7
23 .
Question 16
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−2y= 2
Solution
Step 1: Begin by solving the first equation for x.
2x+ 3y= 7
2x= 7 −3y
x=7−3y
2
14
Step 2: Substitute the expression for xinto the second equation and solve
for y.
47−3y
2−2y= 2
14 −6y−2y= 2
14 −8y= 2
−8y=−12
y=−12
−8
y=3
2
Step 3: Substitute the value of yback into the expression for xto find the
value of x.
x=7−3(3
2)
2
x=7−9
2
2
x=14 −9
4
x=5
2
Therefore, the solution to the system of equations is x=5
2and y=3
2.
Question 17
Question
Solve the following system of equations:
(2x+ 3y= 7
3x−5y= 1
Solution
Step 1: Multiply the first equation by 3 and the second equation by 2 to make
the coefficients of xthe same in both equations.
(6x+ 9y= 21
6x−10y= 2
Step 2: Subtract the second equation from the first equation to eliminate x.
(6x+ 9y)−(6x−10y) = 21 −2
19y= 19
y= 1
15
Step 3: Substitute the value of yback into one of the original equations to
solve for x.2x+ 3(1) = 7
2x+ 3 = 7
2x= 4
x= 2
Therefore, the solution to the system of equations is x= 2 and y= 1.
Question 18
Question
Solve the following system of equations:
(3x+ 4y= 5
6x+ 8y= 10
Solution
Step 1: We can start by multiplying the first equation by 2 to make the coeffi-
cients of xthe same so that we can eliminate xwhen we subtract the equations.
Multiply the first equation by 2:
(6x+ 8y= 10
6x+ 8y= 10
Step 2: Subtract the equations to eliminate x.
Subtract the equations:
(3x+ 4y= 5
−(6x+ 8y= 10)
(3x+ 4y= 5
−6x−8y=−10
⇒(−3x−4y=−5
−6x−8y=−10
Step 3: Now, we can multiply the second equation by −1
2to make the
coefficients of xthe same to eliminate xagain.
Multiply the second equation by −1
2:
16
(−3x−4y=−5
(−1
2)(−6x−8y) = (−1
2)(−10)
(−3x−4y=−5
3x+ 4y= 5
Step 4: Add the equations to eliminate x.
Add the equations:
(−3x−4y=−5
+(3x+ 4y= 5)
(−3x−4y=−5
+3x+ 4y= 5
⇒(0 = 0
0 = 0
Step 5: Since we obtained 0 = 0 for both equations, this means the two
equations are equivalent and are satisfied by any pair of values for xand y.
Therefore, the system of equations has infinitely many solutions.
Question 19
Question
Solve the following system of equations:
3x+ 2y−z= 6
2x−y+ 3z=−10
x+ 3y+ 2z= 10
Solution
Step 1: Start by writing the augmented matrix for the system of equations.
3 2 −1|6
2−1 3 | −10
1 3 2 |10
Step 2: Perform row operations to simplify the matrix.
3 2 −1|6
2−1 3 | −10
1 3 2 |10
→
1 3 2 |10
2−1 3 | −10
3 2 −1|6
17
Step 3: Use row operations to eliminate the coefficients below the leading 1
in the first column.
1 3 2 |10
2−1 3 | −10
3 2 −1|6
→
1 3 2 |10
0−7−1| −30
0−7−7| −24
Step 4: Continue to simplify the matrix by performing row operations.
1 3 2 |10
0−7−1| −30
0−7−7| −24
→
1 3 2 |10
0 1 1 |4
0−7−1| −30
Step 5: Use row operations to eliminate the coefficients above and below the
leading 1 in the second column.
1 3 2 |10
0 1 1 |4
0−7−1| −30
→
1 0 −1| −2
0 1 1 |4
0 0 6 |6
Step 6: Divide the third row by 6 to obtain the values of z.
1 0 −1| −2
0 1 1 |4
0 0 1 |1
Step 7: Substitute z= 1 back into the second row to solve for y.
y+ 1 = 4 ⇒y= 3
Step 8: Substitute y= 3 and z= 1 back into the first row to solve for x.
x−3−1 = −2⇒x= 0
Therefore, the solution to the system of equations is x= 0, y= 3, z= 1.
Question 20
Question
Solve the following system of equations:
3x+ 2y−z= 5
4x−y+ 2z=−1
x+ 3y−4z= 6
18
Solution
Step 1: Let’s start by writing the system of equations in matrix form as Ax=b,
where Ais the coefficient matrix, xis the variable matrix, and bis the constant
matrix.
A=
3 2 −1
4−1 2
1 3 −4
,x=
x
y
z
,b=
5
−1
6
Step 2: Now, we will augment matrix Awith matrix band row reduce the
augmented matrix to row-echelon form.
3 2 −1 5
4−1 2 −1
1 3 −4 6
1 3 −4 6
0−10 14 −25
0−7 1 −1
Step 3: Next, we will continue row operations to get the matrix in reduced
row-echelon form, where the system of equations is easily solvable.
1 3 −4 6
0 1 −7
10
5
2
0 0 −39
10 −23
2
Step 4: From the reduced row-echelon form, we can see that the equations
simplify to:
x+ 3y−4z= 6
y−7
10 z=5
2
−39
10 z=−23
2
Solving the system of equations gives:
z=23
39
y=11
3
x=44
39
Therefore, the solution to the system of equations is x=44
39, y =11
3, z =23
39 .
19
Question 21
Question
Solve the following system of equations:
(4x+ 3y= 7
2x+ 5y= 11
Solution
Step 1: Let’s multiply the first equation by 2 and the second equation by 4 to
make the coefficients of xthe same:
(8x+ 6y= 14
8x+ 20y= 44
Step 2: Now, subtract the first equation from the second to eliminate x:
(8x+ 6y= 14
−6y= 30
Step 3: Solve the equation −6y= 30 for y:
y=−5
Step 4: Substitute y=−5 back into the first original equation to find x:
4x+ 3(−5) = 7 =⇒4x−15 = 7 =⇒4x= 22 =⇒x=22
4= 5.5
Therefore, the solution to the system of equations is x= 5.5 and y=−5.
Question 22
Question
Solve the following system of equations:
4x−2y+z= 3
2x+ 3y−z= 6
x−y+ 2z= 1
20
Solution
Step 1: We will start by setting up an augmented matrix for the system of
equations.
4−2 1 3
2 3 −1 6
1−1 2 1
Step 2: Perform row operations to get the augmented matrix into reduced
row-echelon form.
4−2 1 3
2 3 −1 6
1−1 2 1
R1↔R3
−→
1−1 2 1
2 3 −1 6
4−2 1 3
R2−2R1→R2
−→
1−1 2 1
0 5 −5 4
4−2 1 3
R3−4R1→R3
−→
1−1 2 1
0 5 −5 4
0 2 −7−1
Step 3: Continue row operations to further simplify the augmented matrix.
1−1 2 1
0 5 −5 4
0 2 −7−1
R2−2R3→R2
−→
1−1 2 1
0 1 3 6
0 2 −7−1
R3−2R2→R3
−→
1−1 2 1
0 1 3 6
0 0 −13 −13
Step 4: Convert the augmented matrix back into a system of equations.
x−y+ 2z= 1
y+ 3z= 6
−13z=−13
Step 5: Solve the system of equations. From the third equation, we have
z= 1. Substituting z= 1 into the second equation gives y+ 3(1) = 6 ⇒y= 3.
Finally, substituting z= 1 and y= 3 into the first equation gives x−3 + 2(1) =
1⇒x= 2.
Therefore, the solution to the system of equations is x= 2, y= 3, z= 1.
21
Question 23
Question
Solve the following system of equations:
2x+y+z= 10
x−y+ 3z= 4
3x+ 2y−z= 6
Solution
1. We will solve this system of equations using the method of substitution.
2. Consider the first and second equations. We will solve for yin the first
equation and substitute it into the second equation.
3. Step 1: Solve for y in the first equation.
2x+y+z= 10 =⇒y= 10 −2x−z
4. Step 2: Substitute y= 10 −2x−zinto the second equation.
x−(10 −2x−z)+3z= 4
Simplify:
x−10 + 2x+z+ 3z= 4 =⇒3x+ 4z= 14
5. Now we have a new equation in terms of xand z.
6. Consider the third equation in the original system and substitute y=
10 −2x−zinto it.
7. Step 3: Substitute y= 10 −2x−zinto the third equation.
3x+ 2(10 −2x−z)−z= 6
Simplify:
3x+ 20 −4x−2z−z= 6 =⇒ −x−3z=−14
8. Now we have two new equations: 3x+ 4z= 14 and −x−3z=−14.
9. Step 4: Solve the system of equations 3x+ 4z= 14 and −x−3z=
−14.Multiply the second equation by 3 to match the coefficient of x:
−3x−9z=−42
Add the above equation to the first equation:
3x+ 4z+ (−3x−9z) = 14 −42 =⇒ −5z=−28 =⇒z= 5.6
22
10. Step 5: Substitute z= 5.6back into one of the equations to solve
for x.Let’s use the first equation:
2x+y+ 5.6 = 10 =⇒2x+y= 4.4
Since we know y= 10 −2x−z, substitute z= 5.6 into y:
2x−10 + 2x+ 5.6=4.4 =⇒4x−4.4 = 4.4 =⇒x= 2.2
11. Step 6: Finally, find yusing y= 10 −2x−z.
y= 10 −2(2.2) −5.6 = 10 −4.4−5.6=0
12. Therefore, the solution to the system of equations is x= 2.2, y= 0, and
z= 5.6.
Question 24
Question
Consider the following system of equations:
(2x−3y= 7
4x+ 5y= 1
Determine whether the system has one solution, infinitely many solutions,
or no solution. If it has a unique solution, find the values of xand y.
Solution
Step 1: Multiply the first equation by 4 and the second equation by 2 to elimi-
nate y:
(8x−12y= 28
8x+ 10y= 2
Step 2: Subtract the second equation from the first equation:
8x−12y−(8x+ 10y) = 28 −2
−22y= 26
y=−13
11
Step 3: Substitute the value of yback into the first equation to solve for x:
2x−3(−13
11)=7
23
2x+39
11 = 7
2x= 7 −39
11
2x=77
11 −39
11
2x=38
11
x=19
11
Therefore, the system has a unique solution: x=19
11 and y=−13
11 .
Question 25
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 1
Solution
Step 1: Start by solving the second equation for yin terms of x.
4x−y= 1
y= 4x−1
Step 2: Substitute y= 4x−1 into the first equation and solve for x.
2x+ 3(4x−1) = 7
2x+ 12x−3 = 7
14x= 10
x=10
14
x=5
7
Step 3: Substitute x=5
7back into y= 4x−1 to find the value of y.
y= 4 5
7−1
y=20
7−1
y=20
7−7
7
y=13
7
24
Therefore, the solution to the system of equations is x=5
7and y=13
7.
Question 26
Question
Solve the system of equations:
(3x+ 4y= 11
6x−8y= 10
Solution
Step 1: Let’s start by multiplying the first equation by 2 to make the coefficients
of xequal in both equations. Step 2: Multiply both sides of the first equation
by 2: Step 3: The system becomes:
(6x+ 8y= 22
6x−8y= 10
Step 4: Add both equations together to eliminate y: Step 5: (6x+ 8y) + (6x−
8y) = 22 + 10 Step 6: 12x= 32 Step 7: Divide by 12 to solve for x: Step 8:
x=32
12 Step 9: x=8
3Step 10: Substitute x=8
3back into the first equation
to solve for y: Step 11: 3(8
3)+4y= 11 Step 12: 8 + 4y= 11 Step 13: 4y= 3
Step 14: Divide by 4 to solve for y: Step 15: y=3
4Step 16: The solution to
the system of equations is (x, y)=(8
3,3
4).
Question 27
Question
Solve the following system of equations:
(2x−3y= 7
4x+ 5y= 1
Solution
Step 1: Multiply the first equation by 2:
(4x−6y= 14
4x+ 5y= 1
25
Step 2: Subtract the first equation from the second equation to eliminate
4x:(4x+ 5y)−(4x−6y) = 1 −14
11y=−13
y=−13
11
Step 3: Substitute y=−13
11 back into the first equation:
2x−3(−13
11)=7
2x+39
11 = 7
2x=67
11
x=67
22
Therefore, the solution to the system of equations is x=67
22 and y=−13
11 .
Question 28
Question
Solve the following system of equations:
(2x+ 3y= 5
4x−2y= 6
Solution
Step 1: Let’s solve the first equation for xin terms of y.
2x+ 3y= 5 =⇒2x= 5 −3y=⇒x=5−3y
2
Step 2: Substitute xin the second equation, and solve for y.
45−3y
2−2y= 6 =⇒10−6y−2y= 6 =⇒10−8y= 6 =⇒ −8y=−4 =⇒y=−4
−8=1
2
Step 3: Substitute yback into the equation solved for x, and solve for x.
x=5−3(1
2)
2=5−3
2
2=10 −3
4=7
4
Therefore, the solution to the system of equations is x=7
4and y=1
2.
26
Question 29
Question
Find the values of xand ythat satisfy the following system of equations:
(2x−3y= 7
3x+ 2y=−1
Solution
Step 1: First, we will solve this system of equations by elimination.
Step 2: We will multiply the first equation by 3 and the second equation by
2 to make the yterms cancel when added together.
(6x−9y= 21
6x+ 4y=−2
Step 3: Now, subtract the second equation from the first equation to elimi-
nate the xterms.
−13y= 23
Step 4: Divide by −13 to solve for y.
y=−23
13
Step 5: Substitute y=−23
13 back into one of the original equations to solve
for x. Let’s use the first equation.
2x−3−23
13= 7
Step 6: Simplify and solve for x.
2x+69
13 = 7
2x= 7 −69
13
2x=91
13
x=91
26
Step 7: Therefore, the solution to the system of equations is x=91
26 and
y=−23
13 .
27
Question 30
Question
Solve the following system of equations:
(2x+ 3y= 7
3x−y= 10
Solution
Step 1: Solve the second equation for yin terms of x.
3x−y= 10
y= 3x−10
Step 2: Substitute y= 3x−10 into the first equation and solve for x.
2x+ 3(3x−10) = 7
2x+ 9x−30 = 7
11x= 37
x=37
11 = 3.36
Step 3: Substitute x= 3.36 back into y= 3x−10 to solve for y.
y= 3(3.36) −10
y= 10.08 −10
y= 0.083
Hence, the solution to the system of equations is x=37
11 and y= 0.083.
Question 31
Question
Solve the following system of equations:
(2x+ 3y= 8
4x−5y=−7
Solution
Step 1: Begin by multiplying the first equation by 5 and the second equation
by 3 to make the coefficients of yin each equation equal.
(10x+ 15y= 40
12x−15y=−21
28
Step 2: Add the two equations together to eliminate y.
22x= 19
Step 3: Solve for xby dividing both sides of the equation by 22.
x=19
22
Step 4: Substitute x=19
22 back into one of the original equations. Let’s use
the first equation 2x+ 3y= 8.
219
22+ 3y= 8
Step 5: Simplify the equation and solve for y.
38
22 + 3y= 8
38 + 66y
22 = 8
38 + 66y= 8 ×22
66y= 176 −38
66y= 138
y= 2
Step 6: The solution to the system of equations is x=19
22 and y= 2.
Question 32
Question
Solve the following system of equations:
(2x−y= 5
x2+y2= 29
Solution
Step 1: Begin by solving the first equation for yin terms of x.
2x−y= 5
y= 2x−5
29
Step 2: Substitute yin terms of xinto the second equation to solve for x.
x2+ (2x−5)2= 29
x2+ 4x2−20x+ 25 = 29
5x2−20x−4=0
x2−4x−1=0
Step 3: Solve the quadratic equation x2−4x−1 = 0 using the quadratic
formula:
x=−(−4) ±p(−4)2−4(1)(−1)
2(1) x=4±√16 + 4
2x=4±√20
2x=4±2√5
2x= 2±√5
Step 4: Substituting x= 2 + √5 and y= 2x−5 into the first equation to
find y.
y= 2(2 + √5) −5
y= 4 + 2√5−5
y= 2√5−1
Step 5: Therefore, the solution to the system of equations is:
(x= 2 + √5
y= 2√5−1
Question 33
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y2=−1
Solution
Step 1: Let’s start by solving the first equation for xin terms of y.
Step 1: 2x+ 3y= 7 =⇒x=7−3y
2
Step 2: Substitute xin the second equation with the expression we found in
Step 1.
Step 2: 4 7−3y
2−y2=−1
30
Step 3: Simplify the equation obtained in Step 2.
Step 3: 14 −6y−y2=−1
Step 4: Rearrange the equation to form a quadratic equation.
Step 4: y2+ 6y−15 = 0
Step 5: Solve the quadratic equation by factoring:
Step 5: (y+ 5)(y−3) = 0 =⇒y=−5, y = 3
Step 6: Substitute the values of yback into the equation obtained in Step 1
to find the corresponding values of x.
Step 6: (y=−5 =⇒x=7−3(−5)
2=⇒x= 8
y= 3 =⇒x=7−3(3)
2=⇒x=−1
Therefore, the solutions to the system of equations are (x, y) = (8,−5) and
(x, y)=(−1,3).
Question 34
Question
Solve the following system of equations:
(2x+ 3y= 13
3x−4y= 4
Solution
Step 1: Multiply the first equation by 3 and the second equation by 2 to create
opposites for y.
(6x+ 9y= 39
6x−8y= 8
Step 2: Subtract the second equation from the first equation to eliminate x.
17y= 31
Step 3: Solve for y.
y=31
17
Step 4: Substitute yback into one of the original equations to solve for x.
2x+ 3 31
17= 13
31
Step 5: Solve for x.
x=23
17
Therefore, the solution to the system of equations is:
x=23
17, y =31
17
Question 35
Question
Solve the following system of equations:
(3x−2y= 1
x2+ 2xy −3y2=−6
Solution
Step 1: Rearrange the equations so that the second equation is in terms of one
variable.
x2+ 2xy −3y2=−6
x2+ 2xy −3y2+ 6 = 0
(x+y)(x−3y) + 6 = 0
Step 2: Substitute (x+y) with 1
3from the first equation to solve for y.
1
3(x) + 6 = 0
x+ 18 = 0
x=−18
Step 3: Substitute x=−18 back into the first equation to solve for y.
3(−18) −2y= 1
−54 −2y= 1
−2y= 55
y=−55
2
Step 4: Check the solution by substituting x=−18 and y=−55
2into the
second equation.
(−18)2+ 2(−18)(−55
2)−3(55
2)2=−6
324 + 594 + 907.5 = −6
1800.5=−6
32
Solve for x:
2x−15
11 = 1
2x= 1 + 15
11 =11
11 +15
11 =26
11
x=13
11
Step 5: Therefore, the solution to the system of equations is:
(x=13
11
y=5
11
Question 2
Question
Solve the following system of equations:
(3x−y= 5
x2+y2= 25
Solution
Step 1: Solve the first equation for yin terms of x.
3x−y= 5 =⇒y= 3x−5
Step 2: Substitute y= 3x−5 into the second equation.
x2+ (3x−5)2= 25
Step 3: Expand and simplify the equation.
x2+ 9x2−30x+ 25 = 25
10x2−30x= 0
10x(x−3) = 0
Step 4: Solve for xby setting each factor to zero.
10x= 0 =⇒x= 0
x−3 = 0 =⇒x= 3
Step 5: Now, substitute the values of xback into the equation y= 3x−5
to find the corresponding values of y. When x= 0:
y= 3(0) −5 = −5 =⇒(x, y) = (0,−5)
When x= 3:
y= 3(3) −5 = 4 =⇒(x, y) = (3,4)
Step 6: The solution to the system of equations is (0,−5) and (3,4).
2
Question 3
Question
Solve the following system of equations:
(3x+ 5y= 17
2x−4y= 6
Solution
Step 1: We can solve this system of equations using the method of elimination.
Let’s start by multiplying the first equation by 2 and the second equation by 3
to create oppositely signed coefficients for x.
Step 2:
(6x+ 10y= 34
6x−12y= 18
Step 3: Now, we subtract the second equation from the first equation to
eliminate x:(6x+ 10y)−(6x−12y) = 34 −18
22y= 16
y=16
22
y=8
11
Step 4: Next, we substitute the value of yback into the first equation to
solve for x:
3x+ 5( 8
11) = 17
3x+40
11 = 17
3x= 17 −40
11
3x=187
11
x=187
33
Step 5: Therefore, the solution to the system of equations is:
(x=187
33
y=8
11
3
Question 4
Question
Solve the following system of equations:
(3x+ 2y= 7
4x+ 3y= 9
Solution
Step 1: We can solve the system of equations using the method of substitution.
Let’s solve the first equation for x: From the first equation 3x+ 2y= 7, we have
3x= 7 −2y
x=7−2y
3
Step 2: Now, substitute xin terms of yinto the second equation. Substitute
x=7−2y
3into the second equation 4x+ 3y= 9:
47−2y
3+ 3y= 9
Step 3: Simplify the equation:
28 −8y
3+ 3y= 9
28 −8y+ 9y= 27
y=−1
Step 4: Substitute y=−1 back into x=7−2y
3to find x:
x=7−2(−1)
3
x=9
3
x= 3
Step 5: Therefore, the solution to the system of equations is x= 3 and
y=−1.
Question 5
Question
Solve the following system of equations:
(2x−3y= 5
4x+y= 1
4
Solution
Step 1: Start by solving the second equation for yin terms of x.
4x+y= 1
y=−4x+ 1
Step 2: Substitute y=−4x+ 1 into the first equation to solve for x.
2x−3(−4x+ 1) = 5
2x+ 12x−3=5
14x−3=5
14x= 8
x=8
14
x=4
7
Step 3: Substitute x=4
7back into y=−4x+ 1 to find y.
y=−44
7+ 1
y=−16
7+ 1
y=−16
7+7
7
y=−9
7
Therefore, the solution to the system of equations is (x, y) = 4
7,−9
7.
Question 6
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 5
Solution
Step 1: Let’s solve the second equation for y.
4x−y= 5
−y= 5 −4x
y=−5+4x
5
Step 2: Substitute yfrom the second equation into the first equation.
2x+ 3(−5+4x)=7
2x−15 + 12x= 7
14x−15 = 7
14x= 22
x=22
14
x=11
7
Step 3: Now, substitute xback into the second equation to solve for y.
411
7−y= 5
44
7−y= 5
−y= 5 −44
7
−y=35 −44
7
−y=−9
7
y=9
7
Step 4: Therefore, the solution to the system of equations is x=11
7and
y=9
7.
Question 7
Question
Solve the following system of equations:
2x−3y+z= 7
x+ 2y−2z= 1
3x−5y−3z= 4
Solution
Step 1: We will begin by writing the system of equations in matrix form. Let
A=
2−3 1
1 2 −2
3−5−3
6
and
X=
x
y
z
,
where Xis the column matrix containing the variables x,y, and z, and rewrite
the system as AX=B, where B=
7
1
4
.
Step 2: Next, we will find the inverse of matrix Aby row reducing the
augmented matrix [A|I]:
2−3 1 |100
1 2 −2|010
3−5−3|001
Row reducing gives us
1 0 −5| −3 2 6
0 1 −4|2−1 3
0 0 0 |13 −5−9
Since the augmented matrix is not in reduced row-echelon form, the system
of equations is inconsistent, meaning there is no solution.
Question 8
Question
Solve the following system of equations:
2x−y+ 3z= 4
x+ 2y−z= 1
3x−2y+ 2z= 7
Solution
Step 1: Let’s first simplify the system by writing it as an augmented matrix and
then perform row operations to put it in row-echelon form.
2−1 3 |4
1 2 −1|1
3−2 2 |7
Step 2: We’ll start by subtracting 0.5×row 1 from row 2, and 1.5×row 1
from row 3.
2−1 3 |4
0 2.5−2.5| −1
0−3.5−2.5|1
7
Step 3: Next, we’ll multiply row 2 by 2
5to simplify calculations, and then
add 1.4×row 2 to row 3.
2−1 3 |4
0 1 −1| −0.4
0 0 −6|0.6
Step 4: We’ll divide row 3 by −6 to get the reduced row-echelon form.
2−1 3 |4
0 1 −1| −0.4
0 0 1 | −0.1
Step 5: Now, we will back-substitute to find the values of x,y, and z. From
the third row, z=−0.1. Substitute zinto the second row: y−1 = −0.4⇒y=
0.6. Substitute yand zinto the first row: 2x+ 0.6 + 0.3=4⇒2x= 3.1⇒x=
1.55.
Therefore, the solution to the system of equations is x= 1.55, y= 0.6, and
z=−0.1.
Question 9
Question
Find the solution to the following system of equations:
(2x+ 3y= 11
3x−2y= 1
Solution
Step 1: Let’s solve the first equation for xin terms of y.
2x+ 3y= 11
2x= 11 −3y
x=11 −3y
2
8
Step 2: Substitute xin the second equation.
311 −3y
2−2y= 1
33 −9y
2−2y= 1
33 −9y−4y= 2
33 −13y= 2
−13y=−31
y=31
13
y= 3
Step 3: Substitute y= 3 back into the first equation to find x.
2x+ 3(3) = 11
2x+ 9 = 11
2x= 2
x= 1
Thus, the solution to the system of equations is x= 1 and y= 3.
Question 10
Question
Solve the following system of equations:
(3x+ 2y= 7
2x−y= 1
Solution
Step 1: Solve the second equation for y:
2x−y= 1
y= 2x−1
Step 2: Substitute y= 2x−1 into the first equation and solve for x:
3x+ 2(2x−1) = 7
3x+ 4x−2=7
7x= 9
x=9
7
9
Step 3: Substitute x=9
7back into y= 2x−1 to solve for y:
y= 2 9
7−1
y=18
7−1
y=18
7−7
7
y=11
7
Therefore, the solution to the system of equations is x=9
7and y=11
7.
Question 11
Question
Solve the system of equations:
(2x−3y=−1
3x+ 4y= 9
Solution
Step 1: Multiply both sides of the first equation by 3 and the second equation
by 2 in order to eliminate ywhen adding the two equations together.
(6x−9y=−3
6x+ 8y= 18
Step 2: Subtract the first equation from the second equation to eliminate
x.
17y= 21
Step 3: Solve for yby dividing both sides by 17.
y=21
17
Step 4: Substitute the value of yback into one of the original equations to
solve for x. Using the first equation:
2x−321
17=−1
Step 5: Solve for x.
2x−63
17 =−1
10
2x=−1 + 63
17
2x=−17 + 63
17
2x=−80
17
x=−40
17
Step 6: The solution to the system of equations is x=−40
17 and y=21
17 .
Question 12
Question
Solve the following system of equations:
(3x+ 2y= 5
x2+y2= 10
Solution
Step 1: Let’s start by solving the first equation for yin terms of x.
3x+ 2y= 5
2y= 5 −3x
y=5−3x
2
Step 2: Substitute yby 5−3x
2in the second equation.
x2+5−3x
22
= 10
x2+(5 −3x)2
4= 10
x2+25 −30x+ 9x2
4= 10
4x2+ 25 −30x+ 9x2= 40
13x2−30x−15 = 0
11
Step 3: Solve the quadratic equation 13x2−30x−15 = 0.
x=−(−30) ±p(−30)2−4(13)(−15)
2(13)
x=30 ±√900 + 780
26
x=30 ±√1680
26
x=30 ±4√105
26
x=15 ±2√105
13
Step 4: Substitute the values of xback into the equation y=5−3x
2to find
the corresponding yvalues. Therefore, the solutions to the system of equations
are: 15 + 2√105
13 ,5−3(15+2√105
13 )
2!
and 15 −2√105
13 ,5−3(15−2√105
13 )
2!
Question 13
Question
Consider the following system of equations:
(2x−3y= 5
4x+ky = 11
Find the value(s) of kthat will make this system of equations have a unique
solution.
Solution
Step 1: To find the value(s) of kthat will make this system have a unique
solution, we need to ensure that the two equations represent distinct lines. This
means the lines must not be parallel or coincide.
Step 2: Let’s first write the second equation in slope-intercept form by solv-
ing for y:
4x+ky = 11 =⇒ky =−4x+ 11 =⇒y=−4
kx+11
k
12
Step 3: For the system of equations to have a unique solution, the slopes of
the two lines formed by the equations must not be equal. Therefore, the slope
of the first line 2x−3y= 5 and the slope of the second line −4
kx+11
kshould
not be equal.
Step 4: The slope of the first line 2x−3y= 5 is 2
3. So, for a unique solution
to exist, the slope of the second line −4
kshould not be equal to 2
3. Thus, −4
k=2
3.
Step 5: Solving −4
k=2
3for k, we get:
−4
k=2
3=⇒k=−6
Step 6: Therefore, for the system of equations to have a unique solution, k
must not be equal to −6. Thus, the value of kthat will make this system have
a unique solution is k=−6.
Question 14
Question
Solve the following system of equations:
(4x+ 3y= 7
2x−y= 3
Solution
Step 1: Solve the second equation for y:
2x−y= 3
y= 2x−3
Step 2: Substitute y= 2x−3 into the first equation and solve for x:
4x+ 3(2x−3) = 7
4x+ 6x−9=7
10x−9=7
10x= 16
x=16
10
x= 1.6
Step 3: Substitute x= 1.6 into y= 2x−3 to find y:
y= 2(1.6) −3
y= 3.2−3
y= 0.2
Therefore, the solution to the system of equations is x= 1.6 and y= 0.2.
13
Question 15
Question
Solve the following system of equations for xand y:
(3x+ 4y= 5
5x−y= 7
Solution
Step 1: Multiply the second equation by 4 to eliminate y:
(3x+ 4y= 5
20x−4y= 28
Step 2: Add the equations together to eliminate y:
23x= 33 ⇒x=33
23
Step 3: Substitute xback into the first equation to solve for y:
333
23+ 4y= 5 ⇒y=−7
23
Step 4: The solution to the system of equations is x=33
23 and y=−7
23 .
Question 16
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−2y= 2
Solution
Step 1: Begin by solving the first equation for x.
2x+ 3y= 7
2x= 7 −3y
x=7−3y
2
14
Step 2: Substitute the expression for xinto the second equation and solve
for y.
47−3y
2−2y= 2
14 −6y−2y= 2
14 −8y= 2
−8y=−12
y=−12
−8
y=3
2
Step 3: Substitute the value of yback into the expression for xto find the
value of x.
x=7−3(3
2)
2
x=7−9
2
2
x=14 −9
4
x=5
2
Therefore, the solution to the system of equations is x=5
2and y=3
2.
Question 17
Question
Solve the following system of equations:
(2x+ 3y= 7
3x−5y= 1
Solution
Step 1: Multiply the first equation by 3 and the second equation by 2 to make
the coefficients of xthe same in both equations.
(6x+ 9y= 21
6x−10y= 2
Step 2: Subtract the second equation from the first equation to eliminate x.
(6x+ 9y)−(6x−10y) = 21 −2
19y= 19
y= 1
15
Step 3: Substitute the value of yback into one of the original equations to
solve for x.2x+ 3(1) = 7
2x+ 3 = 7
2x= 4
x= 2
Therefore, the solution to the system of equations is x= 2 and y= 1.
Question 18
Question
Solve the following system of equations:
(3x+ 4y= 5
6x+ 8y= 10
Solution
Step 1: We can start by multiplying the first equation by 2 to make the coeffi-
cients of xthe same so that we can eliminate xwhen we subtract the equations.
Multiply the first equation by 2:
(6x+ 8y= 10
6x+ 8y= 10
Step 2: Subtract the equations to eliminate x.
Subtract the equations:
(3x+ 4y= 5
−(6x+ 8y= 10)
(3x+ 4y= 5
−6x−8y=−10
⇒(−3x−4y=−5
−6x−8y=−10
Step 3: Now, we can multiply the second equation by −1
2to make the
coefficients of xthe same to eliminate xagain.
Multiply the second equation by −1
2:
16
(−3x−4y=−5
(−1
2)(−6x−8y) = (−1
2)(−10)
(−3x−4y=−5
3x+ 4y= 5
Step 4: Add the equations to eliminate x.
Add the equations:
(−3x−4y=−5
+(3x+ 4y= 5)
(−3x−4y=−5
+3x+ 4y= 5
⇒(0 = 0
0 = 0
Step 5: Since we obtained 0 = 0 for both equations, this means the two
equations are equivalent and are satisfied by any pair of values for xand y.
Therefore, the system of equations has infinitely many solutions.
Question 19
Question
Solve the following system of equations:
3x+ 2y−z= 6
2x−y+ 3z=−10
x+ 3y+ 2z= 10
Solution
Step 1: Start by writing the augmented matrix for the system of equations.
3 2 −1|6
2−1 3 | −10
1 3 2 |10
Step 2: Perform row operations to simplify the matrix.
3 2 −1|6
2−1 3 | −10
1 3 2 |10
→
1 3 2 |10
2−1 3 | −10
3 2 −1|6
17
Step 3: Use row operations to eliminate the coefficients below the leading 1
in the first column.
1 3 2 |10
2−1 3 | −10
3 2 −1|6
→
1 3 2 |10
0−7−1| −30
0−7−7| −24
Step 4: Continue to simplify the matrix by performing row operations.
1 3 2 |10
0−7−1| −30
0−7−7| −24
→
1 3 2 |10
0 1 1 |4
0−7−1| −30
Step 5: Use row operations to eliminate the coefficients above and below the
leading 1 in the second column.
1 3 2 |10
0 1 1 |4
0−7−1| −30
→
1 0 −1| −2
0 1 1 |4
0 0 6 |6
Step 6: Divide the third row by 6 to obtain the values of z.
1 0 −1| −2
0 1 1 |4
0 0 1 |1
Step 7: Substitute z= 1 back into the second row to solve for y.
y+ 1 = 4 ⇒y= 3
Step 8: Substitute y= 3 and z= 1 back into the first row to solve for x.
x−3−1 = −2⇒x= 0
Therefore, the solution to the system of equations is x= 0, y= 3, z= 1.
Question 20
Question
Solve the following system of equations:
3x+ 2y−z= 5
4x−y+ 2z=−1
x+ 3y−4z= 6
18
Solution
Step 1: Let’s start by writing the system of equations in matrix form as Ax=b,
where Ais the coefficient matrix, xis the variable matrix, and bis the constant
matrix.
A=
3 2 −1
4−1 2
1 3 −4
,x=
x
y
z
,b=
5
−1
6
Step 2: Now, we will augment matrix Awith matrix band row reduce the
augmented matrix to row-echelon form.
3 2 −1 5
4−1 2 −1
1 3 −4 6
1 3 −4 6
0−10 14 −25
0−7 1 −1
Step 3: Next, we will continue row operations to get the matrix in reduced
row-echelon form, where the system of equations is easily solvable.
1 3 −4 6
0 1 −7
10
5
2
0 0 −39
10 −23
2
Step 4: From the reduced row-echelon form, we can see that the equations
simplify to:
x+ 3y−4z= 6
y−7
10 z=5
2
−39
10 z=−23
2
Solving the system of equations gives:
z=23
39
y=11
3
x=44
39
Therefore, the solution to the system of equations is x=44
39, y =11
3, z =23
39 .
19
Question 21
Question
Solve the following system of equations:
(4x+ 3y= 7
2x+ 5y= 11
Solution
Step 1: Let’s multiply the first equation by 2 and the second equation by 4 to
make the coefficients of xthe same:
(8x+ 6y= 14
8x+ 20y= 44
Step 2: Now, subtract the first equation from the second to eliminate x:
(8x+ 6y= 14
−6y= 30
Step 3: Solve the equation −6y= 30 for y:
y=−5
Step 4: Substitute y=−5 back into the first original equation to find x:
4x+ 3(−5) = 7 =⇒4x−15 = 7 =⇒4x= 22 =⇒x=22
4= 5.5
Therefore, the solution to the system of equations is x= 5.5 and y=−5.
Question 22
Question
Solve the following system of equations:
4x−2y+z= 3
2x+ 3y−z= 6
x−y+ 2z= 1
20
Solution
Step 1: We will start by setting up an augmented matrix for the system of
equations.
4−2 1 3
2 3 −1 6
1−1 2 1
Step 2: Perform row operations to get the augmented matrix into reduced
row-echelon form.
4−2 1 3
2 3 −1 6
1−1 2 1
R1↔R3
−→
1−1 2 1
2 3 −1 6
4−2 1 3
R2−2R1→R2
−→
1−1 2 1
0 5 −5 4
4−2 1 3
R3−4R1→R3
−→
1−1 2 1
0 5 −5 4
0 2 −7−1
Step 3: Continue row operations to further simplify the augmented matrix.
1−1 2 1
0 5 −5 4
0 2 −7−1
R2−2R3→R2
−→
1−1 2 1
0 1 3 6
0 2 −7−1
R3−2R2→R3
−→
1−1 2 1
0 1 3 6
0 0 −13 −13
Step 4: Convert the augmented matrix back into a system of equations.
x−y+ 2z= 1
y+ 3z= 6
−13z=−13
Step 5: Solve the system of equations. From the third equation, we have
z= 1. Substituting z= 1 into the second equation gives y+ 3(1) = 6 ⇒y= 3.
Finally, substituting z= 1 and y= 3 into the first equation gives x−3 + 2(1) =
1⇒x= 2.
Therefore, the solution to the system of equations is x= 2, y= 3, z= 1.
21
Question 23
Question
Solve the following system of equations:
2x+y+z= 10
x−y+ 3z= 4
3x+ 2y−z= 6
Solution
1. We will solve this system of equations using the method of substitution.
2. Consider the first and second equations. We will solve for yin the first
equation and substitute it into the second equation.
3. Step 1: Solve for y in the first equation.
2x+y+z= 10 =⇒y= 10 −2x−z
4. Step 2: Substitute y= 10 −2x−zinto the second equation.
x−(10 −2x−z)+3z= 4
Simplify:
x−10 + 2x+z+ 3z= 4 =⇒3x+ 4z= 14
5. Now we have a new equation in terms of xand z.
6. Consider the third equation in the original system and substitute y=
10 −2x−zinto it.
7. Step 3: Substitute y= 10 −2x−zinto the third equation.
3x+ 2(10 −2x−z)−z= 6
Simplify:
3x+ 20 −4x−2z−z= 6 =⇒ −x−3z=−14
8. Now we have two new equations: 3x+ 4z= 14 and −x−3z=−14.
9. Step 4: Solve the system of equations 3x+ 4z= 14 and −x−3z=
−14.Multiply the second equation by 3 to match the coefficient of x:
−3x−9z=−42
Add the above equation to the first equation:
3x+ 4z+ (−3x−9z) = 14 −42 =⇒ −5z=−28 =⇒z= 5.6
22
10. Step 5: Substitute z= 5.6back into one of the equations to solve
for x.Let’s use the first equation:
2x+y+ 5.6 = 10 =⇒2x+y= 4.4
Since we know y= 10 −2x−z, substitute z= 5.6 into y:
2x−10 + 2x+ 5.6=4.4 =⇒4x−4.4 = 4.4 =⇒x= 2.2
11. Step 6: Finally, find yusing y= 10 −2x−z.
y= 10 −2(2.2) −5.6 = 10 −4.4−5.6=0
12. Therefore, the solution to the system of equations is x= 2.2, y= 0, and
z= 5.6.
Question 24
Question
Consider the following system of equations:
(2x−3y= 7
4x+ 5y= 1
Determine whether the system has one solution, infinitely many solutions,
or no solution. If it has a unique solution, find the values of xand y.
Solution
Step 1: Multiply the first equation by 4 and the second equation by 2 to elimi-
nate y:
(8x−12y= 28
8x+ 10y= 2
Step 2: Subtract the second equation from the first equation:
8x−12y−(8x+ 10y) = 28 −2
−22y= 26
y=−13
11
Step 3: Substitute the value of yback into the first equation to solve for x:
2x−3(−13
11)=7
23
2x+39
11 = 7
2x= 7 −39
11
2x=77
11 −39
11
2x=38
11
x=19
11
Therefore, the system has a unique solution: x=19
11 and y=−13
11 .
Question 25
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 1
Solution
Step 1: Start by solving the second equation for yin terms of x.
4x−y= 1
y= 4x−1
Step 2: Substitute y= 4x−1 into the first equation and solve for x.
2x+ 3(4x−1) = 7
2x+ 12x−3 = 7
14x= 10
x=10
14
x=5
7
Step 3: Substitute x=5
7back into y= 4x−1 to find the value of y.
y= 4 5
7−1
y=20
7−1
y=20
7−7
7
y=13
7
24
Therefore, the solution to the system of equations is x=5
7and y=13
7.
Question 26
Question
Solve the system of equations:
(3x+ 4y= 11
6x−8y= 10
Solution
Step 1: Let’s start by multiplying the first equation by 2 to make the coefficients
of xequal in both equations. Step 2: Multiply both sides of the first equation
by 2: Step 3: The system becomes:
(6x+ 8y= 22
6x−8y= 10
Step 4: Add both equations together to eliminate y: Step 5: (6x+ 8y) + (6x−
8y) = 22 + 10 Step 6: 12x= 32 Step 7: Divide by 12 to solve for x: Step 8:
x=32
12 Step 9: x=8
3Step 10: Substitute x=8
3back into the first equation
to solve for y: Step 11: 3(8
3)+4y= 11 Step 12: 8 + 4y= 11 Step 13: 4y= 3
Step 14: Divide by 4 to solve for y: Step 15: y=3
4Step 16: The solution to
the system of equations is (x, y)=(8
3,3
4).
Question 27
Question
Solve the following system of equations:
(2x−3y= 7
4x+ 5y= 1
Solution
Step 1: Multiply the first equation by 2:
(4x−6y= 14
4x+ 5y= 1
25
Step 2: Subtract the first equation from the second equation to eliminate
4x:(4x+ 5y)−(4x−6y) = 1 −14
11y=−13
y=−13
11
Step 3: Substitute y=−13
11 back into the first equation:
2x−3(−13
11)=7
2x+39
11 = 7
2x=67
11
x=67
22
Therefore, the solution to the system of equations is x=67
22 and y=−13
11 .
Question 28
Question
Solve the following system of equations:
(2x+ 3y= 5
4x−2y= 6
Solution
Step 1: Let’s solve the first equation for xin terms of y.
2x+ 3y= 5 =⇒2x= 5 −3y=⇒x=5−3y
2
Step 2: Substitute xin the second equation, and solve for y.
45−3y
2−2y= 6 =⇒10−6y−2y= 6 =⇒10−8y= 6 =⇒ −8y=−4 =⇒y=−4
−8=1
2
Step 3: Substitute yback into the equation solved for x, and solve for x.
x=5−3(1
2)
2=5−3
2
2=10 −3
4=7
4
Therefore, the solution to the system of equations is x=7
4and y=1
2.
26
Question 29
Question
Find the values of xand ythat satisfy the following system of equations:
(2x−3y= 7
3x+ 2y=−1
Solution
Step 1: First, we will solve this system of equations by elimination.
Step 2: We will multiply the first equation by 3 and the second equation by
2 to make the yterms cancel when added together.
(6x−9y= 21
6x+ 4y=−2
Step 3: Now, subtract the second equation from the first equation to elimi-
nate the xterms.
−13y= 23
Step 4: Divide by −13 to solve for y.
y=−23
13
Step 5: Substitute y=−23
13 back into one of the original equations to solve
for x. Let’s use the first equation.
2x−3−23
13= 7
Step 6: Simplify and solve for x.
2x+69
13 = 7
2x= 7 −69
13
2x=91
13
x=91
26
Step 7: Therefore, the solution to the system of equations is x=91
26 and
y=−23
13 .
27
Question 30
Question
Solve the following system of equations:
(2x+ 3y= 7
3x−y= 10
Solution
Step 1: Solve the second equation for yin terms of x.
3x−y= 10
y= 3x−10
Step 2: Substitute y= 3x−10 into the first equation and solve for x.
2x+ 3(3x−10) = 7
2x+ 9x−30 = 7
11x= 37
x=37
11 = 3.36
Step 3: Substitute x= 3.36 back into y= 3x−10 to solve for y.
y= 3(3.36) −10
y= 10.08 −10
y= 0.083
Hence, the solution to the system of equations is x=37
11 and y= 0.083.
Question 31
Question
Solve the following system of equations:
(2x+ 3y= 8
4x−5y=−7
Solution
Step 1: Begin by multiplying the first equation by 5 and the second equation
by 3 to make the coefficients of yin each equation equal.
(10x+ 15y= 40
12x−15y=−21
28
Step 2: Add the two equations together to eliminate y.
22x= 19
Step 3: Solve for xby dividing both sides of the equation by 22.
x=19
22
Step 4: Substitute x=19
22 back into one of the original equations. Let’s use
the first equation 2x+ 3y= 8.
219
22+ 3y= 8
Step 5: Simplify the equation and solve for y.
38
22 + 3y= 8
38 + 66y
22 = 8
38 + 66y= 8 ×22
66y= 176 −38
66y= 138
y= 2
Step 6: The solution to the system of equations is x=19
22 and y= 2.
Question 32
Question
Solve the following system of equations:
(2x−y= 5
x2+y2= 29
Solution
Step 1: Begin by solving the first equation for yin terms of x.
2x−y= 5
y= 2x−5
29
Step 2: Substitute yin terms of xinto the second equation to solve for x.
x2+ (2x−5)2= 29
x2+ 4x2−20x+ 25 = 29
5x2−20x−4=0
x2−4x−1=0
Step 3: Solve the quadratic equation x2−4x−1 = 0 using the quadratic
formula:
x=−(−4) ±p(−4)2−4(1)(−1)
2(1) x=4±√16 + 4
2x=4±√20
2x=4±2√5
2x= 2±√5
Step 4: Substituting x= 2 + √5 and y= 2x−5 into the first equation to
find y.
y= 2(2 + √5) −5
y= 4 + 2√5−5
y= 2√5−1
Step 5: Therefore, the solution to the system of equations is:
(x= 2 + √5
y= 2√5−1
Question 33
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y2=−1
Solution
Step 1: Let’s start by solving the first equation for xin terms of y.
Step 1: 2x+ 3y= 7 =⇒x=7−3y
2
Step 2: Substitute xin the second equation with the expression we found in
Step 1.
Step 2: 4 7−3y
2−y2=−1
30
Step 3: Simplify the equation obtained in Step 2.
Step 3: 14 −6y−y2=−1
Step 4: Rearrange the equation to form a quadratic equation.
Step 4: y2+ 6y−15 = 0
Step 5: Solve the quadratic equation by factoring:
Step 5: (y+ 5)(y−3) = 0 =⇒y=−5, y = 3
Step 6: Substitute the values of yback into the equation obtained in Step 1
to find the corresponding values of x.
Step 6: (y=−5 =⇒x=7−3(−5)
2=⇒x= 8
y= 3 =⇒x=7−3(3)
2=⇒x=−1
Therefore, the solutions to the system of equations are (x, y) = (8,−5) and
(x, y)=(−1,3).
Question 34
Question
Solve the following system of equations:
(2x+ 3y= 13
3x−4y= 4
Solution
Step 1: Multiply the first equation by 3 and the second equation by 2 to create
opposites for y.
(6x+ 9y= 39
6x−8y= 8
Step 2: Subtract the second equation from the first equation to eliminate x.
17y= 31
Step 3: Solve for y.
y=31
17
Step 4: Substitute yback into one of the original equations to solve for x.
2x+ 3 31
17= 13
31
Step 5: Solve for x.
x=23
17
Therefore, the solution to the system of equations is:
x=23
17, y =31
17
Question 35
Question
Solve the following system of equations:
(3x−2y= 1
x2+ 2xy −3y2=−6
Solution
Step 1: Rearrange the equations so that the second equation is in terms of one
variable.
x2+ 2xy −3y2=−6
x2+ 2xy −3y2+ 6 = 0
(x+y)(x−3y) + 6 = 0
Step 2: Substitute (x+y) with 1
3from the first equation to solve for y.
1
3(x) + 6 = 0
x+ 18 = 0
x=−18
Step 3: Substitute x=−18 back into the first equation to solve for y.
3(−18) −2y= 1
−54 −2y= 1
−2y= 55
y=−55
2
Step 4: Check the solution by substituting x=−18 and y=−55
2into the
second equation.
(−18)2+ 2(−18)(−55
2)−3(55
2)2=−6
324 + 594 + 907.5 = −6
1800.5=−6
32
The solution x=−18, y =−55
2does not satisfy the second equation, so
there is no solution for this system of equations.
33
