MATH 114 - QUANTITATIVE
REASONING - Systems of Equations
Question Bank - Set 5
Liberty University
Question 1
Question
Find all solutions to the following system of equations:
(3x−2y= 5
2x+y= 1
Solution
Step 1: Start by solving the second equation for y.
2x+y= 1
y= 1 −2x
Step 2: Substitute y= 1 −2xinto the first equation and solve for x.
3x−2(1 −2x) = 5
3x−2+4x= 5
7x−2 = 5
7x= 7
x= 1
Step 3: Substitute x= 1 back into the second equation to find y.
2(1) + y= 1
2 + y= 1
y=−1
Therefore, the solution to the system of equations is x= 1 and y=−1.
Question 2
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 5
Solution
Step 1: Let’s solve the second equation for yin terms of x.
4x−y= 5
y= 4x−5
Step 2: Substitute y= 4x−5 into the first equation.
2x+ 3(4x−5) = 7
2x+ 12x−15 = 7
14x−15 = 7
Step 3: Now, solve for x.
14x−15 = 7
14x= 22
x=22
14
x=11
7
Step 4: Substitute x=11
7back into y= 4x−5 to find y.
y= 4 11
7−5
y=44
7−35
7
y=9
7
Therefore, the solution to the system of equations is x=11
7and y=9
7.
Question 3
Question
Solve the following system of equations:
(2x−3y= 8
4x+ 5y= 7
2
Solution
Step 1: To solve this system of equations, we will use the method of elimination.
We will multiply the first equation by 5 and the second equation by 3 in order
to eliminate ywhen adding the equations together.
Step 2: Multiplying the first equation by 5:
5(2x−3y)=5·8
Simplifying, we get:
10x−15y= 40
Step 3: Multiplying the second equation by 3:
3(4x+ 5y)=3·7
Simplifying, we get:
12x+ 15y= 21
Step 4: Adding the two new equations together:
10x−15y+ 12x+ 15y= 40 + 21
Simplifying, we get:
22x= 61
Step 5: Solving for x:
x=61
22
Step 6: Substituting xback into the first equation to find y:
261
22−3y= 8
122
22 −3y= 8
122
22 −66
22 = 3y
56
22 = 3y
y=28
33
Step 7: Therefore, the solution to the system of equations is:
x=61
22, y =28
33
3
Question 4
Question
Solve the following system of equations:
(3x−2y= 5
4x+y= 3
Solution
Step 1: Start by solving the second equation for yin terms of x.
4x+y= 3
y= 3 −4x
Step 2: Substitute y= 3 −4xinto the first equation and solve for x.
3x−2(3 −4x)=5
3x−6+8x= 5
11x= 11
x= 1
Step 3: Substitute x= 1 back into the second equation to solve for y.
4(1) + y= 3
4 + y= 3
y=−1
Step 4: The solution to the system is x= 1 and y=−1. Therefore, the
solution to the system of equations is (1,−1).
Question 5
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 8
Solution
Step 1: Start by solving the second equation for y.
4x−y= 8
y= 4x−8
4
Step 2: Substitute y= 4x−8 into the first equation.
2x+ 3(4x−8) = 7
2x+ 12x−24 = 7
14x−24 = 7
14x= 31
x=31
14
Step 3: Substitute x=31
14 back into y= 4x−8 to find y.
y= 4 31
14−8
y=31
7−8
y=31 −56
7
y=−25
7
Therefore, the solution to the system of equations is x=31
14 and y=−25
7.
Question 6
Question
Solve the following system of equations:
2x+y−z= 4
x−y+z= 2
x+ 2y+ 3z= 7
Solution
Step 1: Begin by rewriting the system of equations in matrix form:
2 1 −1
1−1 1
1 2 3
x
y
z
=
4
2
7
Step 2: Perform row operations to get the matrix into reduced row-echelon
form:
100
010
001
x
y
z
=
3
1
2
Step 3: The solution to the system of equations is x= 3, y= 1, and z= 2.
5
Question 7
Question
Solve the following system of equations:
(2x−3y= 5
x2+ 4y2= 17
Solution
Let’s solve the system of equations step by step.
Step 1: Begin by isolating one variable in the first equation. We will solve
for xin terms of y.
2x−3y= 5
2x= 3y+ 5
x=3y+ 5
2
Step 2: Substitute this expression for xinto the second equation.
3y+ 5
22
+ 4y2= 17
Step 3: Expand and simplify the equation.
9y2+ 30y+ 25
4+ 4y2= 17
9y2+ 30y+ 25 + 16y2= 68
25y2+ 30y−43 = 0
Step 4: Solve the quadratic equation for y.
y=−30 ±√302−4·25 · −43
2·25
y=−30 ±√900 + 4300
50
y=−30 ±√5200
50
y=−30 ±2√1300
50
y=−15 ±√1300
25
Step 5: Calculate the values of yand then find the corresponding values
of xusing x=3y+5
2. Thus, we have two possible solutions for the system of
equations:
Solution 1: (x, y) = 1
2,−2and Solution 2: (x, y) = 9
2,2
5
6
Question 8
Question
Solve the following system of equations:
2x−y+ 3z= 7
x+y−z= 2
3x+ 2y+z= 5
Solution
Step 1: Let’s write the system of equations in matrix form:
2−1 3
1 1 −1
3 2 1
x
y
z
=
7
2
5
Step 2: We will now row-reduce the augmented matrix:
2−137
1 1 −1 2
3 2 1 5
Step 3: Let’s perform row operations to simplify the matrix. First, let’s
multiply the second row by 2 and subtract the first row multiplied by 1 to
eliminate the xterm.
0 3 −5−3
1 1 −1 2
3 2 1 5
Step 4: Next, let’s multiply the third row by 2 and subtract 3 times the
second row to eliminate the xterm in the third row.
0 3 −5−3
1 1 −1 2
0−1 5 −1
Step 5: To simplify further, let’s multiply the third row by 3 and add the
second row to eliminate the yterm in the third row.
0 3 −5−3
1 1 −1 2
0 0 14 5
Step 6: Finally, divide the third row by 14 to solve for z. The other variables
can then be solved by back substitution.
So, the solution to the system of equations is x=−1, y = 2, z =5
14 .
7
Question 9
Question
Solve the following system of equations:
(2x+ 3y= 11
4x−y=−3
Solution
Step 1: Solve the second equation for yin terms of x.
4x−y=−3
y= 4x+ 3
Step 2: Substitute y= 4x+ 3 into the first equation.
2x+ 3(4x+ 3) = 11
2x+ 12x+ 9 = 11
14x+ 9 = 11
14x= 2
x=2
14
x=1
7
Step 3: Substitute x=1
7back into the equation y= 4x+ 3 to find y.
y= 4 1
7+ 3
y=4
7+ 3
y=4
7+21
7
y=25
7
Step 4: The solution to the system of equations is x=1
7and y=25
7.
Question 10
Question
Solve the following system of equations using the elimination method:
(3x+ 2y= 13
4x−5y=−9
8
Solution
Step 1: Multiply the first equation by 5 and the second equation by 2 to elimi-
nate y:
(15x+ 10y= 65
8x−10y=−18
Step 2: Add the equations together to solve for x:
15x+ 10y+ 8x−10y= 65 −18
23x= 47
x= 2
Step 3: Substitute x= 2 into the first equation and solve for y:
3(2) + 2y= 13
6+2y= 13
2y= 7
y=7
2
Therefore, the solution to the system of equations is x= 2 and y=7
2.
Question 11
Question
Solve the following system of equations:
(2x+ 3y= 5
4x−y= 7
Solution
Step 1: Solve the second equation for yin terms of x:
4x−y= 7 =⇒y= 4x−7
Step 2: Substitute y= 4x−7 into the first equation and solve for x:
2x+ 3(4x−7) = 5
2x+ 12x−21 = 5
14x−21 = 5
14x= 26
9
x=26
14 =13
7
Step 3: Substitute x=13
7back into y= 4x−7 to find y:
y= 4 13
7−7 = 52
7−7 = 52 −49
7=3
7
Therefore, the solution to the system of equations is x=13
7and y=3
7.
Question 12
Question
Solve the following system of equations:
2x+y+z= 6
x−y+ 2z=−1
3x+ 2y−z= 7
Solution
Step 1: Rewrite the system of equations in matrix form and compute the aug-
mented matrix.
2 1 1 |6
1−1 2 | −1
3 2 −1|7
Step 2: Perform row operations to echelon form.
1−1 2 | −1
0 3/2−3/2|7/2
0 5 −7|20
Step 3: Perform row operations to reduced echelon form.
1 0 1 |2
0 1 −1|5
0 0 0 |0
Step 4: Write the system of equations in parametric form.
(x= 2 −z
y= 5 + z
Thus, the solution to the system of equations is x= 2 −z,y= 5 + z,zis a
free parameter.
10
Question 13
Question
Solve the following system of equations:
4x−y+ 3z= 2
2x+y−z= 4
−x+ 2y+z=−3
Solution
Step 1: Let’s label the equations for easier reference:
(1) 4x−y+ 3z= 2
(2) 2x+y−z= 4
(3) −x+ 2y+z=−3
Step 2: We will use the method of elimination to solve the system. Let’s
start by adding equations (1) and (3) to eliminate y:
4x−y+ 3z= 2
−x+ 2y+z=−3
3x+ 4z=−1
Step 3: Next, let’s add equations (2) and (3) to eliminate y:
2x+y−z= 4
−x+ 2y+z=−3
x+z= 1
Step 4: We now have a new system of equations:
((4) 3x+ 4z=−1
(5) x+z= 1
Step 5: Now, let’s solve equations (4) and (5) simultaneously. First, we will
multiply equation (5) by 3 and subtract it from equation (4) to eliminate x:
3(3x+ 4z) = 3(−1)
−3x−3z=−3
3(x+z)=3
z= 0
Step 6: Substituting z= 0 back into equation (5), we find:
x+ 0 = 1 =⇒x= 1
11
Step 7: Finally, substituting x= 1 and z= 0 back into equation (2), we can
solve for y:
2(1) + y−0 = 4 =⇒y= 2
Therefore, the solution to the system of equations is x= 1, y= 2, and z= 0.
Question 14
Question
Solve the following system of equations:
(2x−3y= 7
4x+ 5y= 6
Solution
Step 1: Multiply the first equation by 2 to eliminate xwhen adding the two
equations together.
(4x−6y= 14
4x+ 5y= 6
Step 2: Subtract the modified first equation from the second equation to
eliminate x.
(4x−6y= 14
11y=−8
Step 3: Solve for yby dividing both sides of the equation by 11.
y=−8
11
Step 4: Substitute the value of yback into one of the original equations.
Let’s use the first equation.
2x−3−8
11 = 7
Step 5: Simplify and solve for x.
2x+24
11 = 7
2x=77
11 −24
11
2x=53
11
12
x=53
22
Step 6: The solution to the system of equations is x=53
22 and y=−8
11 .
Question 15
Question
Solve the following system of equations:
(2x+ 3y= 12
4x+ 6y= 24
Solution
Step 1: We can see that the second equation is a multiple of the first equation.
We can simplify the system by dividing the second equation by 2:
(2x+ 3y= 12
2x+ 3y= 12
Step 2: We can see that both equations are the same. This means that the
system has infinitely many solutions. The solution set can be described in terms
of xor y, such as x=tand y=12−2t
3where tis any real number.
Question 16
Question
Solve the following system of equations:
2x−3y+ 5z= 1
x+ 2y−z= 4
3x−2y+ 4z=−3
Solution
Step 1: Let’s write the system of equations in matrix form AX =B, where Ais
the coefficient matrix, Xis the unknown matrix, and Bis the constant matrix:
2−3 5
1 2 −1
3−2 4
x
y
z
=
1
4
−3
13
Step 2: To solve for X, we need to find the inverse of matrix Asince X=
A−1B.
Step 3: Let’s calculate the determinant of matrix A:
det(A) = det
2−3 5
1 2 −1
3−2 4
det(A) = 2(2 ×4 + 1 ×(−2)) −(−3)(1 ×4 + 3 ×(−1)) + 5(1 ×−2−2×3) = 25
Step 4: Now, let’s find the inverse of matrix A:
A−1=1
det(A)adj(A)
Where adj(A) is the adjugate of matrix A.
Step 5: Calculate the adjugate of matrix A:
adj(A) = cof(A)T
Step 6: By finding the cofactor matrix of A, we get:
cof(A) =
12 −9 7
−5 14 −10
−4 7 −4
Step 7: Transposing the cofactor matrix gives us the adjugate:
adj(A) =
12 −5−4
−9 14 7
7−10 −4
Step 8: Now, we can find the inverse of matrix A:
A−1=1
25
12 −5−4
−9 14 7
7−10 −4
Step 9: Finally, we can solve for X:
X=A−1B=1
25
12 −5−4
−9 14 7
7−10 −4
1
4
−3
After performing the matrix multiplication, we can obtain the values of x,
y, and z.
14
Question 17
Question
Solve the following system of equations:
(3x+ 2y= 11
x−y= 2
Solution
Step 1: Rearrange the second equation for xin terms of y.
x−y= 2
x=y+ 2
Step 2: Substitute xin the first equation with y+ 2.
3(y+ 2) + 2y= 11
3y+ 6 + 2y= 11
5y+ 6 = 11
5y= 5
y= 1
Step 3: Substitute yback into x=y+ 2 to find x.
x= 1 + 2
x= 3
Therefore, the solution to the system of equations is x= 3 and y= 1.
Question 18
Question
Solve the following system of equations:
(3x+ 2y= 7
x2−y2= 1
Solution
Step 1: Solve the first equation for x:
3x+ 2y= 7
3x= 7 −2y
x=7−2y
3
15
Step 2: Substitute the expression for xinto the second equation:
7−2y
32
−y2= 1
(7 −2y)2
9−y2= 1
Step 3: Expand the expression and simplify:
49 −28y+ 4y2
9−y2= 1
49 −28y+ 4y2−9y2
9= 1
4y2−9y2−28y+ 49
9= 1
−5y2−28y+ 49
9= 1
Step 4: Multiply through by 9 to clear the fraction:
−5y2−28y+ 49 = 9
Step 5: Rearrange the equation into a quadratic form:
−5y2−28y+ 40 = 0
Step 6: Solve the quadratic equation through factoring, completing the
square, or using the quadratic formula. Let’s use the quadratic formula:
y=−(−28) ±p(−28)2−4(−5)(40)
2(−5)
y=28 ±√784 + 800
−10
y=28 ±√1584
−10
y=28 ±2√396
−10
y=28 ±2·2√99
−10
y=28 ±4√99
−10
y=28 ±4√9√11
−10
y=28 ±4·3√11
−10
y=28 ±12√11
−10
16
Therefore, the solutions for yare y=28+12√11
−10 and y=28−12√11
−10 . Substi-
tuting these values back into x=7−2y
3will give the corresponding values for
x.
Question 19
Question
Solve the following system of equations:
(x2+y2= 25
x+y= 5
Solution
Step 1: We can solve the second equation for one variable and substitute it into
the first equation. Solving for xin x+y= 5 gives x= 5 −y. Substitute this
into x2+y2= 25:
(5 −y)2+y2= 25
Step 2: Expand and simplify the equation:
25 −10y+y2+y2= 25
2y2−10y= 0
2y(y−5) = 0
Step 3: Set each factor to zero and solve for y:
2y= 0 =⇒y= 0
y−5 = 0 =⇒y= 5
Step 4: For y= 0, substitute back into x= 5 −yto find x:
x= 5 −0
x= 5
Step 5: For y= 5, substitute back into x= 5 −yto find x:
x= 5 −5
x= 0
Step 6: The solutions to the system are x= 5, y= 0 and x= 0, y= 5.
17
Question 20
Question
Solve the following system of equations:
(2x−3y= 7
3x+ 2y= 8
Solution
Step 1: Let’s start by solving the first equation for xin terms of y:
2x−3y= 7 =⇒2x= 3y+ 7 =⇒x=3y+ 7
2
Step 2: Now, substitute x=3y+7
2into the second equation:
33y+ 7
2+ 2y= 8
Step 3: Simplifying, we get:
9y+ 21
2+ 2y= 8
9y+ 21 + 4y= 16
13y+ 21 = 16
13y=−5
y=−5
13
Step 4: Substitute y=−5
13 back into x=3y+7
2to find x:
x=3(−5
13 )+7
2
x=−15/13 + 91/13
2
x=76/13
2
x=38
13
Step 5: Therefore, the solution to the system of equations is:
x=38
13 and y=−5
13
18
Question 21
Question
Solve the following system of equations:
(3x−y= 5
x2+y2= 13
Solution
Step 1: Solve the first equation for yin terms of x.
3x−y= 5
y= 3x−5
Step 2: Substitute yin terms of xinto the second equation.
x2+ (3x−5)2= 13
x2+ 9x2−30x+ 25 = 13
10x2−30x+ 12 = 0
5x2−15x+ 6 = 0
Step 3: Solve the quadratic equation for x.
x=−(−15) ±p(−15)2−4(5)(6)
2(5)
x=15 ±√225 −120
10
x=15 ±√105
10
x=15 ±√3·52
10
x=15 ±5√3
10
x=3±√3
2(simplified)
19
Step 4: Find the corresponding values of y.
y= 3 3±√3
2!−5
y=9±3√3
2−5
y=9±3√3−10
2
y=−1±3√3
2
Therefore, the solution to the system of equations is:
(x=3+√3
2, y =−1+3√3
2
x=3−√3
2, y =−1−3√3
2
Question 22
Question
Solve the following system of equations:
(2x+ 3y= 7
3x−2y= 1
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 and solve for y.
37−3y
2−2y= 1 =⇒21 −9y
2−2y= 1
21 −9y−4y
2= 1 =⇒21 −13y
2= 1 =⇒21−13y= 2 =⇒ −13y=−19 =⇒y=19
13
Step 3: Substitute yback into the equation solved for xto find the value of
x.
x=7−3(19
13 )
2=7−57
13
2=91 −57
26 =34
26 =17
13
Therefore, the solution to the system of equations is x=17
13 and y=19
13 .
20
Question 23
Question
Solve the following system of equations:
(2x+ 3y= 8
4x−y= 5
Solution
Step 1: Solve the second equation for yin terms of x.
4x−y= 5
y= 4x−5
Step 2: Substitute y= 4x−5 into the first equation.
2x+ 3(4x−5) = 8
2x+ 12x−15 = 8
14x−15 = 8
14x= 23
x=23
14
Step 3: Substitute x=23
14 back into the equation y= 4x−5 to find y.
y= 4 23
14−5
y=92
14 −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 24
Question
Solve the following system of equations:
(3x−2y= 4
x2+y2= 10
21
Solution
Step 1: Solve the first equation for xin terms of y.
3x−2y= 4
3x= 2y+ 4
x=2y+ 4
3
Step 2: Substitute the expression for xinto the second equation.
2y+ 4
32
+y2= 10
4y2+ 16y+ 16
9+y2= 10
4y2+ 16y+ 16 + 9y2= 90
13y2+ 16y+ 16 = 90
13y2+ 16y−74 = 0
Step 3: Solve the quadratic equation for y.
y=−16 ±p162−4(13)(−74)
2(13)
y=−16 ±√256 + 3832
26
y=−16 ±√4088
26
y=−16 ±64
26
Thus, we have y=48
26 =24
13 or y=−80
26 =−40
13 .
Step 4: Find the corresponding values of xfor each value of yobtained in
Step 3. For y=24
13 :
x=224
13 + 4
3
x=48/13 + 4
3
x=48/13 + 52/13
3
x=100/13
3
x=100
39
22
For y=−40
13 :
x=2−40
13 + 4
3
x=−80/13 + 4
3
x=−80/13 + 52/13
3
x=−28/13
3
x=−28
39
Therefore, the solutions to the system of equations are x=100
39 and y=24
13
or x=−28
39 and y=−40
13 .
Question 25
Question
Solve the following system of equations:
(2x+ 3y= 13
4x−y= 3
Solution
Step 1: Rearrange the second equation to solve for y:
4x−y= 3
−y=−4x+ 3
y= 4x−3
Step 2: Substitute y= 4x−3 into the first equation:
2x+ 3(4x−3) = 13
2x+ 12x−9 = 13
14x= 22
x=22
14
x=11
7
23
Step 3: Substitute x=11
7back into y= 4x−3 to solve for y:
y= 4 11
7−3
y=44
7−3
y=44
7−21
7
y=23
7
Therefore, the solution to the system of equations is x=11
7and y=23
7.
Question 26
Question
Find the solution to the following system of equations:
3x+ 2y−z= 4
x−y+ 3z=−3
2x+y+z= 6
Solution
Step 1: Rearrange the equations into a more convenient form for elimination.
We will eliminate zfirst.
3x+ 2y−z= 4 (Equation 1)
x−y+ 3z=−3 (Equation 2)
2x+y+z= 6 (Equation 3)
Step 2: Add Equations 2 and 3 to eliminate z.
(x−y+ 3z) + (2x+y+z)=(−3) + 6
3x+ 2x= 3
5x= 3
x=3
5
24
Step 3: Substitute x=3
5into Equation 1 to solve for y.
33
5+ 2y−z= 4
9
5+ 2y−z= 4
2y−z= 1 −9
5
2y−z=−4
5
Step 4: Substitute x=3
5into Equation 3 to solve for z.
23
5+y+z= 6
6
5+y+z= 6
y+z= 6 −6
5
y+z=24
5
Step 5: Substitute y+z=24
5into 2y−z=−4
5to solve for y.
2y−24
5=−4
5
2y=24
5−4
5
2y=20
5
y= 2
Therefore, the solution to the system of equations is x=3
5,y= 2, and
z=14
5.
Question 27
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 5
25
Solution
Step 1: Let’s solve the second equation for y:
4x−y= 5 =⇒y= 4x−5
Step 2: Substitute the expression for yinto the first equation:
2x+ 3(4x−5) = 7
Step 3: Simplify and solve for x:
2x+ 12x−15 = 714x−15 = 714x= 22x=22
14 =11
7
Step 4: Substitute the value of xback into the equation y= 4x−5 to find
y:
y= 4 11
7−5 = 44
7−35
7=9
7
Therefore, the solution to the system of equations is x=11
7and y=9
7.
Question 28
Question
Solve the following system of equations:
(2x+ 3y= 7
3x−2y= 4
Solution
Step 1: We will use the method of substitution to solve the system of equations.
Step 2: Solve the first equation for x:
2x+ 3y= 7 =⇒x=7−3y
2
Step 3: Substitute this expression for xinto the second equation:
37−3y
2−2y= 4
Step 4: Simplify the equation:
21 −9y
2−2y= 4
Step 5: Multiply through by 2 to clear the fraction:
21 −9y−4y= 8
26
Step 6: Combine like terms:
21 −13y= 8
Step 7: Solve for y:
−13y=−13
Step 8: Divide by -13 to find y:
y= 1
Step 9: Substitute y= 1 back into the equation we found for x:
x=7−3(1)
2=4
2= 2
Step 10: Therefore, the solution to the system of equations is (x, y) = (2,1).
Question 29
Question
Solve the following system of equations:
(3x−2y= 7
2x+y= 4
Solution
Step 1: We will solve the second equation for y:
2x+y= 4
y= 4 −2x
Step 2: Substitute the expression for yinto the first equation:
3x−2(4 −2x) = 7
3x−8+4x= 7
7x−8 = 7
Step 3: Solve the equation for x:
7x−8 = 7
7x= 15
x=15
7
27
Step 4: Substitute the value of xback into y= 4 −2xto find y:
y= 4 −215
7
y= 4 −30
7
y=28
7−30
7
y=−2
7
Therefore, the solution to the system of equations is x=15
7and y=−2
7.
Question 30
Question
Solve the following system of equations:
(2x+ 3y= 10
3x+ 2y= 8
Solution
Step 1: Let’s multiply the first equation by 3 and the second equation by 2 to
create opposite coefficients for xwhen we add the equations together.
(6x+ 9y= 30
6x+ 4y= 16
Step 2: Subtract the second equation from the first equation to eliminate x.
5y= 14
Step 3: Solve for y.
y=14
5
Step 4: Substitute the value of yback into either of the original equations.
Let’s use the first equation.
2x+ 3 14
5= 10
Step 5: Solve for x.
2x+42
5= 10
2x= 10 −42
5
28
2x=8
5
x=4
5
Step 6: The solution to the system of equations is x=4
5and y=14
5.
Question 31
Question
Solve the following system of equations:
(3x+ 2y= 10
2x−3y=−5
Solution
Step 1: We will solve the system of equations using the method of elimination.
First, we will multiply the first equation by 3 and the second equation by 2 in
order to create opposite coefficients of y.
Step 2: Multiplying the first equation by 3, we get:
(9x+ 6y= 30
2x−3y=−5
Step 3: Multiplying the second equation by 2, we get:
(9x+ 6y= 30
4x−6y=−10
Step 4: Now we will add the two equations together to eliminate y:
(9x+ 6y) + (4x−6y) = 30 + (−10)
13x= 20
x=20
13
29
Step 5: Substituting x=20
13 back into the first equation, we can solve for y:
320
13+ 2y= 10
60
13 + 2y= 10
2y= 10 −60
13
2y=130
13 −60
13
2y=70
13
y=35
13
Step 6: Therefore, the solution to the system of equations is:
x=20
13, y =35
13
Question 32
Question
Solve the following system of equations:
(2x−3y= 5
4x+y= 3
Solution
Step 1: Solve the second equation for y:
4x+y= 3 =⇒y= 3 −4x
Step 2: Substitute y= 3 −4xinto the first equation and solve for x:
2x−3(3 −4x) = 5
2x−9 + 12x= 5
14x−9=5
14x= 14
x= 1
Step 3: Substitute x= 1 back into the second equation and solve for y:
4(1) + y= 3
30
4 + y= 3
y= 3 −4
y=−1
Step 4: Verify the solution by substituting x= 1 and y=−1 into both
original equations:
(2(1) −3(−1) = 5 (True)
4(1) + (−1) = 3 (True)
Therefore, the solution to the system of equations is x= 1 and y=−1.
Question 33
Question
Solve the following system of equations:
(2x−5y= 17
3x+ 2y= 1
Solution
Step 1: Let’s begin by multiplying the first equation by 2 and the second equa-
tion by 5 to eliminate y.
Multiplying the first equation by 2:
(4x−10y= 34
3x+ 2y= 1
Multiplying the second equation by 5:
(4x−10y= 34
15x+ 10y= 5
Step 2: Now, we will add the two new equations to eliminate y:
(4x−10y= 34
15x+ 10y= 5
(4x−10y) + (15x+ 10y) = 34 + 5
19x= 39
x=39
19
31
Step 3: Next, we will substitute the value of xback into one of the original
equations to solve for y. Let’s use the first equation 2x−5y= 17:
239
19−5y= 17
78
19 −5y= 17
−5y= 17 −78
19
−5y=323
19 −78
19
−5y=245
19
y=245
−19 ·5
y=−49
19
Step 4: Therefore, the solution to the system of equations is x=39
19 and
y=−49
19 .
Question 34
Question
Solve the following system of equations:
(2x+ 3y= 7
3x−2y= 9
Solution
Step 1: Let’s solve the system of equations by elimination. Multiply the first
equation by 2 and the second equation by 3 to make the coefficients of yequal:
(4x+ 6y= 14
9x−6y= 27
Step 2: Add the modified equations to eliminate y:
13x= 41
Step 3: Solve for x:
x=41
13 = 3.15
32
Step 4: Substitute xback into one of the original equations, for example,
2x+ 3y= 7, to solve for y:
2(3.15) + 3y= 7
6.3+3y= 7
3y= 0.7
y=0.7
3= 0.23
Step 5: The solution to the system of equations is x= 3.15 and y= 0.23.
Question 35
Question
Solve the following system of equations:
(2x+y= 5
3x−2y=−2
Solution
Step 1: We can solve this system of equations using the method of substitution.
Let’s solve the first equation for y:
y= 5 −2x
Step 2: Substitute y= 5 −2xinto the second equation and solve for x:
3x−2(5 −2x) = −2
Step 3: Simplify and solve for x:
3x−10 + 4x=−2
7x−10 = −2
7x= 8
x=8
7
Step 4: Substitute x=8
7back into y= 5 −2xto find y:
y= 5 −28
7
y= 5 −16
7
33
Question 2
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 5
Solution
Step 1: Let’s solve the second equation for yin terms of x.
4x−y= 5
y= 4x−5
Step 2: Substitute y= 4x−5 into the first equation.
2x+ 3(4x−5) = 7
2x+ 12x−15 = 7
14x−15 = 7
Step 3: Now, solve for x.
14x−15 = 7
14x= 22
x=22
14
x=11
7
Step 4: Substitute x=11
7back into y= 4x−5 to find y.
y= 4 11
7−5
y=44
7−35
7
y=9
7
Therefore, the solution to the system of equations is x=11
7and y=9
7.
Question 3
Question
Solve the following system of equations:
(2x−3y= 8
4x+ 5y= 7
2
Solution
Step 1: To solve this system of equations, we will use the method of elimination.
We will multiply the first equation by 5 and the second equation by 3 in order
to eliminate ywhen adding the equations together.
Step 2: Multiplying the first equation by 5:
5(2x−3y)=5·8
Simplifying, we get:
10x−15y= 40
Step 3: Multiplying the second equation by 3:
3(4x+ 5y)=3·7
Simplifying, we get:
12x+ 15y= 21
Step 4: Adding the two new equations together:
10x−15y+ 12x+ 15y= 40 + 21
Simplifying, we get:
22x= 61
Step 5: Solving for x:
x=61
22
Step 6: Substituting xback into the first equation to find y:
261
22−3y= 8
122
22 −3y= 8
122
22 −66
22 = 3y
56
22 = 3y
y=28
33
Step 7: Therefore, the solution to the system of equations is:
x=61
22, y =28
33
3
Question 4
Question
Solve the following system of equations:
(3x−2y= 5
4x+y= 3
Solution
Step 1: Start by solving the second equation for yin terms of x.
4x+y= 3
y= 3 −4x
Step 2: Substitute y= 3 −4xinto the first equation and solve for x.
3x−2(3 −4x)=5
3x−6+8x= 5
11x= 11
x= 1
Step 3: Substitute x= 1 back into the second equation to solve for y.
4(1) + y= 3
4 + y= 3
y=−1
Step 4: The solution to the system is x= 1 and y=−1. Therefore, the
solution to the system of equations is (1,−1).
Question 5
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 8
Solution
Step 1: Start by solving the second equation for y.
4x−y= 8
y= 4x−8
4
Step 2: Substitute y= 4x−8 into the first equation.
2x+ 3(4x−8) = 7
2x+ 12x−24 = 7
14x−24 = 7
14x= 31
x=31
14
Step 3: Substitute x=31
14 back into y= 4x−8 to find y.
y= 4 31
14−8
y=31
7−8
y=31 −56
7
y=−25
7
Therefore, the solution to the system of equations is x=31
14 and y=−25
7.
Question 6
Question
Solve the following system of equations:
2x+y−z= 4
x−y+z= 2
x+ 2y+ 3z= 7
Solution
Step 1: Begin by rewriting the system of equations in matrix form:
2 1 −1
1−1 1
1 2 3
x
y
z
=
4
2
7
Step 2: Perform row operations to get the matrix into reduced row-echelon
form:
100
010
001
x
y
z
=
3
1
2
Step 3: The solution to the system of equations is x= 3, y= 1, and z= 2.
5
Question 7
Question
Solve the following system of equations:
(2x−3y= 5
x2+ 4y2= 17
Solution
Let’s solve the system of equations step by step.
Step 1: Begin by isolating one variable in the first equation. We will solve
for xin terms of y.
2x−3y= 5
2x= 3y+ 5
x=3y+ 5
2
Step 2: Substitute this expression for xinto the second equation.
3y+ 5
22
+ 4y2= 17
Step 3: Expand and simplify the equation.
9y2+ 30y+ 25
4+ 4y2= 17
9y2+ 30y+ 25 + 16y2= 68
25y2+ 30y−43 = 0
Step 4: Solve the quadratic equation for y.
y=−30 ±√302−4·25 · −43
2·25
y=−30 ±√900 + 4300
50
y=−30 ±√5200
50
y=−30 ±2√1300
50
y=−15 ±√1300
25
Step 5: Calculate the values of yand then find the corresponding values
of xusing x=3y+5
2. Thus, we have two possible solutions for the system of
equations:
Solution 1: (x, y) = 1
2,−2and Solution 2: (x, y) = 9
2,2
5
6
Question 8
Question
Solve the following system of equations:
2x−y+ 3z= 7
x+y−z= 2
3x+ 2y+z= 5
Solution
Step 1: Let’s write the system of equations in matrix form:
2−1 3
1 1 −1
3 2 1
x
y
z
=
7
2
5
Step 2: We will now row-reduce the augmented matrix:
2−137
1 1 −1 2
3 2 1 5
Step 3: Let’s perform row operations to simplify the matrix. First, let’s
multiply the second row by 2 and subtract the first row multiplied by 1 to
eliminate the xterm.
0 3 −5−3
1 1 −1 2
3 2 1 5
Step 4: Next, let’s multiply the third row by 2 and subtract 3 times the
second row to eliminate the xterm in the third row.
0 3 −5−3
1 1 −1 2
0−1 5 −1
Step 5: To simplify further, let’s multiply the third row by 3 and add the
second row to eliminate the yterm in the third row.
0 3 −5−3
1 1 −1 2
0 0 14 5
Step 6: Finally, divide the third row by 14 to solve for z. The other variables
can then be solved by back substitution.
So, the solution to the system of equations is x=−1, y = 2, z =5
14 .
7
Question 9
Question
Solve the following system of equations:
(2x+ 3y= 11
4x−y=−3
Solution
Step 1: Solve the second equation for yin terms of x.
4x−y=−3
y= 4x+ 3
Step 2: Substitute y= 4x+ 3 into the first equation.
2x+ 3(4x+ 3) = 11
2x+ 12x+ 9 = 11
14x+ 9 = 11
14x= 2
x=2
14
x=1
7
Step 3: Substitute x=1
7back into the equation y= 4x+ 3 to find y.
y= 4 1
7+ 3
y=4
7+ 3
y=4
7+21
7
y=25
7
Step 4: The solution to the system of equations is x=1
7and y=25
7.
Question 10
Question
Solve the following system of equations using the elimination method:
(3x+ 2y= 13
4x−5y=−9
8
Solution
Step 1: Multiply the first equation by 5 and the second equation by 2 to elimi-
nate y:
(15x+ 10y= 65
8x−10y=−18
Step 2: Add the equations together to solve for x:
15x+ 10y+ 8x−10y= 65 −18
23x= 47
x= 2
Step 3: Substitute x= 2 into the first equation and solve for y:
3(2) + 2y= 13
6+2y= 13
2y= 7
y=7
2
Therefore, the solution to the system of equations is x= 2 and y=7
2.
Question 11
Question
Solve the following system of equations:
(2x+ 3y= 5
4x−y= 7
Solution
Step 1: Solve the second equation for yin terms of x:
4x−y= 7 =⇒y= 4x−7
Step 2: Substitute y= 4x−7 into the first equation and solve for x:
2x+ 3(4x−7) = 5
2x+ 12x−21 = 5
14x−21 = 5
14x= 26
9
x=26
14 =13
7
Step 3: Substitute x=13
7back into y= 4x−7 to find y:
y= 4 13
7−7 = 52
7−7 = 52 −49
7=3
7
Therefore, the solution to the system of equations is x=13
7and y=3
7.
Question 12
Question
Solve the following system of equations:
2x+y+z= 6
x−y+ 2z=−1
3x+ 2y−z= 7
Solution
Step 1: Rewrite the system of equations in matrix form and compute the aug-
mented matrix.
2 1 1 |6
1−1 2 | −1
3 2 −1|7
Step 2: Perform row operations to echelon form.
1−1 2 | −1
0 3/2−3/2|7/2
0 5 −7|20
Step 3: Perform row operations to reduced echelon form.
1 0 1 |2
0 1 −1|5
0 0 0 |0
Step 4: Write the system of equations in parametric form.
(x= 2 −z
y= 5 + z
Thus, the solution to the system of equations is x= 2 −z,y= 5 + z,zis a
free parameter.
10
Question 13
Question
Solve the following system of equations:
4x−y+ 3z= 2
2x+y−z= 4
−x+ 2y+z=−3
Solution
Step 1: Let’s label the equations for easier reference:
(1) 4x−y+ 3z= 2
(2) 2x+y−z= 4
(3) −x+ 2y+z=−3
Step 2: We will use the method of elimination to solve the system. Let’s
start by adding equations (1) and (3) to eliminate y:
4x−y+ 3z= 2
−x+ 2y+z=−3
3x+ 4z=−1
Step 3: Next, let’s add equations (2) and (3) to eliminate y:
2x+y−z= 4
−x+ 2y+z=−3
x+z= 1
Step 4: We now have a new system of equations:
((4) 3x+ 4z=−1
(5) x+z= 1
Step 5: Now, let’s solve equations (4) and (5) simultaneously. First, we will
multiply equation (5) by 3 and subtract it from equation (4) to eliminate x:
3(3x+ 4z) = 3(−1)
−3x−3z=−3
3(x+z)=3
z= 0
Step 6: Substituting z= 0 back into equation (5), we find:
x+ 0 = 1 =⇒x= 1
11
Step 7: Finally, substituting x= 1 and z= 0 back into equation (2), we can
solve for y:
2(1) + y−0 = 4 =⇒y= 2
Therefore, the solution to the system of equations is x= 1, y= 2, and z= 0.
Question 14
Question
Solve the following system of equations:
(2x−3y= 7
4x+ 5y= 6
Solution
Step 1: Multiply the first equation by 2 to eliminate xwhen adding the two
equations together.
(4x−6y= 14
4x+ 5y= 6
Step 2: Subtract the modified first equation from the second equation to
eliminate x.
(4x−6y= 14
11y=−8
Step 3: Solve for yby dividing both sides of the equation by 11.
y=−8
11
Step 4: Substitute the value of yback into one of the original equations.
Let’s use the first equation.
2x−3−8
11 = 7
Step 5: Simplify and solve for x.
2x+24
11 = 7
2x=77
11 −24
11
2x=53
11
12
x=53
22
Step 6: The solution to the system of equations is x=53
22 and y=−8
11 .
Question 15
Question
Solve the following system of equations:
(2x+ 3y= 12
4x+ 6y= 24
Solution
Step 1: We can see that the second equation is a multiple of the first equation.
We can simplify the system by dividing the second equation by 2:
(2x+ 3y= 12
2x+ 3y= 12
Step 2: We can see that both equations are the same. This means that the
system has infinitely many solutions. The solution set can be described in terms
of xor y, such as x=tand y=12−2t
3where tis any real number.
Question 16
Question
Solve the following system of equations:
2x−3y+ 5z= 1
x+ 2y−z= 4
3x−2y+ 4z=−3
Solution
Step 1: Let’s write the system of equations in matrix form AX =B, where Ais
the coefficient matrix, Xis the unknown matrix, and Bis the constant matrix:
2−3 5
1 2 −1
3−2 4
x
y
z
=
1
4
−3
13
Step 2: To solve for X, we need to find the inverse of matrix Asince X=
A−1B.
Step 3: Let’s calculate the determinant of matrix A:
det(A) = det
2−3 5
1 2 −1
3−2 4
det(A) = 2(2 ×4 + 1 ×(−2)) −(−3)(1 ×4 + 3 ×(−1)) + 5(1 ×−2−2×3) = 25
Step 4: Now, let’s find the inverse of matrix A:
A−1=1
det(A)adj(A)
Where adj(A) is the adjugate of matrix A.
Step 5: Calculate the adjugate of matrix A:
adj(A) = cof(A)T
Step 6: By finding the cofactor matrix of A, we get:
cof(A) =
12 −9 7
−5 14 −10
−4 7 −4
Step 7: Transposing the cofactor matrix gives us the adjugate:
adj(A) =
12 −5−4
−9 14 7
7−10 −4
Step 8: Now, we can find the inverse of matrix A:
A−1=1
25
12 −5−4
−9 14 7
7−10 −4
Step 9: Finally, we can solve for X:
X=A−1B=1
25
12 −5−4
−9 14 7
7−10 −4
1
4
−3
After performing the matrix multiplication, we can obtain the values of x,
y, and z.
14
Question 17
Question
Solve the following system of equations:
(3x+ 2y= 11
x−y= 2
Solution
Step 1: Rearrange the second equation for xin terms of y.
x−y= 2
x=y+ 2
Step 2: Substitute xin the first equation with y+ 2.
3(y+ 2) + 2y= 11
3y+ 6 + 2y= 11
5y+ 6 = 11
5y= 5
y= 1
Step 3: Substitute yback into x=y+ 2 to find x.
x= 1 + 2
x= 3
Therefore, the solution to the system of equations is x= 3 and y= 1.
Question 18
Question
Solve the following system of equations:
(3x+ 2y= 7
x2−y2= 1
Solution
Step 1: Solve the first equation for x:
3x+ 2y= 7
3x= 7 −2y
x=7−2y
3
15
Step 2: Substitute the expression for xinto the second equation:
7−2y
32
−y2= 1
(7 −2y)2
9−y2= 1
Step 3: Expand the expression and simplify:
49 −28y+ 4y2
9−y2= 1
49 −28y+ 4y2−9y2
9= 1
4y2−9y2−28y+ 49
9= 1
−5y2−28y+ 49
9= 1
Step 4: Multiply through by 9 to clear the fraction:
−5y2−28y+ 49 = 9
Step 5: Rearrange the equation into a quadratic form:
−5y2−28y+ 40 = 0
Step 6: Solve the quadratic equation through factoring, completing the
square, or using the quadratic formula. Let’s use the quadratic formula:
y=−(−28) ±p(−28)2−4(−5)(40)
2(−5)
y=28 ±√784 + 800
−10
y=28 ±√1584
−10
y=28 ±2√396
−10
y=28 ±2·2√99
−10
y=28 ±4√99
−10
y=28 ±4√9√11
−10
y=28 ±4·3√11
−10
y=28 ±12√11
−10
16
Therefore, the solutions for yare y=28+12√11
−10 and y=28−12√11
−10 . Substi-
tuting these values back into x=7−2y
3will give the corresponding values for
x.
Question 19
Question
Solve the following system of equations:
(x2+y2= 25
x+y= 5
Solution
Step 1: We can solve the second equation for one variable and substitute it into
the first equation. Solving for xin x+y= 5 gives x= 5 −y. Substitute this
into x2+y2= 25:
(5 −y)2+y2= 25
Step 2: Expand and simplify the equation:
25 −10y+y2+y2= 25
2y2−10y= 0
2y(y−5) = 0
Step 3: Set each factor to zero and solve for y:
2y= 0 =⇒y= 0
y−5 = 0 =⇒y= 5
Step 4: For y= 0, substitute back into x= 5 −yto find x:
x= 5 −0
x= 5
Step 5: For y= 5, substitute back into x= 5 −yto find x:
x= 5 −5
x= 0
Step 6: The solutions to the system are x= 5, y= 0 and x= 0, y= 5.
17
Question 20
Question
Solve the following system of equations:
(2x−3y= 7
3x+ 2y= 8
Solution
Step 1: Let’s start by solving the first equation for xin terms of y:
2x−3y= 7 =⇒2x= 3y+ 7 =⇒x=3y+ 7
2
Step 2: Now, substitute x=3y+7
2into the second equation:
33y+ 7
2+ 2y= 8
Step 3: Simplifying, we get:
9y+ 21
2+ 2y= 8
9y+ 21 + 4y= 16
13y+ 21 = 16
13y=−5
y=−5
13
Step 4: Substitute y=−5
13 back into x=3y+7
2to find x:
x=3(−5
13 )+7
2
x=−15/13 + 91/13
2
x=76/13
2
x=38
13
Step 5: Therefore, the solution to the system of equations is:
x=38
13 and y=−5
13
18
Question 21
Question
Solve the following system of equations:
(3x−y= 5
x2+y2= 13
Solution
Step 1: Solve the first equation for yin terms of x.
3x−y= 5
y= 3x−5
Step 2: Substitute yin terms of xinto the second equation.
x2+ (3x−5)2= 13
x2+ 9x2−30x+ 25 = 13
10x2−30x+ 12 = 0
5x2−15x+ 6 = 0
Step 3: Solve the quadratic equation for x.
x=−(−15) ±p(−15)2−4(5)(6)
2(5)
x=15 ±√225 −120
10
x=15 ±√105
10
x=15 ±√3·52
10
x=15 ±5√3
10
x=3±√3
2(simplified)
19
Step 4: Find the corresponding values of y.
y= 3 3±√3
2!−5
y=9±3√3
2−5
y=9±3√3−10
2
y=−1±3√3
2
Therefore, the solution to the system of equations is:
(x=3+√3
2, y =−1+3√3
2
x=3−√3
2, y =−1−3√3
2
Question 22
Question
Solve the following system of equations:
(2x+ 3y= 7
3x−2y= 1
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 and solve for y.
37−3y
2−2y= 1 =⇒21 −9y
2−2y= 1
21 −9y−4y
2= 1 =⇒21 −13y
2= 1 =⇒21−13y= 2 =⇒ −13y=−19 =⇒y=19
13
Step 3: Substitute yback into the equation solved for xto find the value of
x.
x=7−3(19
13 )
2=7−57
13
2=91 −57
26 =34
26 =17
13
Therefore, the solution to the system of equations is x=17
13 and y=19
13 .
20
Question 23
Question
Solve the following system of equations:
(2x+ 3y= 8
4x−y= 5
Solution
Step 1: Solve the second equation for yin terms of x.
4x−y= 5
y= 4x−5
Step 2: Substitute y= 4x−5 into the first equation.
2x+ 3(4x−5) = 8
2x+ 12x−15 = 8
14x−15 = 8
14x= 23
x=23
14
Step 3: Substitute x=23
14 back into the equation y= 4x−5 to find y.
y= 4 23
14−5
y=92
14 −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 24
Question
Solve the following system of equations:
(3x−2y= 4
x2+y2= 10
21
Solution
Step 1: Solve the first equation for xin terms of y.
3x−2y= 4
3x= 2y+ 4
x=2y+ 4
3
Step 2: Substitute the expression for xinto the second equation.
2y+ 4
32
+y2= 10
4y2+ 16y+ 16
9+y2= 10
4y2+ 16y+ 16 + 9y2= 90
13y2+ 16y+ 16 = 90
13y2+ 16y−74 = 0
Step 3: Solve the quadratic equation for y.
y=−16 ±p162−4(13)(−74)
2(13)
y=−16 ±√256 + 3832
26
y=−16 ±√4088
26
y=−16 ±64
26
Thus, we have y=48
26 =24
13 or y=−80
26 =−40
13 .
Step 4: Find the corresponding values of xfor each value of yobtained in
Step 3. For y=24
13 :
x=224
13 + 4
3
x=48/13 + 4
3
x=48/13 + 52/13
3
x=100/13
3
x=100
39
22
For y=−40
13 :
x=2−40
13 + 4
3
x=−80/13 + 4
3
x=−80/13 + 52/13
3
x=−28/13
3
x=−28
39
Therefore, the solutions to the system of equations are x=100
39 and y=24
13
or x=−28
39 and y=−40
13 .
Question 25
Question
Solve the following system of equations:
(2x+ 3y= 13
4x−y= 3
Solution
Step 1: Rearrange the second equation to solve for y:
4x−y= 3
−y=−4x+ 3
y= 4x−3
Step 2: Substitute y= 4x−3 into the first equation:
2x+ 3(4x−3) = 13
2x+ 12x−9 = 13
14x= 22
x=22
14
x=11
7
23
Step 3: Substitute x=11
7back into y= 4x−3 to solve for y:
y= 4 11
7−3
y=44
7−3
y=44
7−21
7
y=23
7
Therefore, the solution to the system of equations is x=11
7and y=23
7.
Question 26
Question
Find the solution to the following system of equations:
3x+ 2y−z= 4
x−y+ 3z=−3
2x+y+z= 6
Solution
Step 1: Rearrange the equations into a more convenient form for elimination.
We will eliminate zfirst.
3x+ 2y−z= 4 (Equation 1)
x−y+ 3z=−3 (Equation 2)
2x+y+z= 6 (Equation 3)
Step 2: Add Equations 2 and 3 to eliminate z.
(x−y+ 3z) + (2x+y+z)=(−3) + 6
3x+ 2x= 3
5x= 3
x=3
5
24
Step 3: Substitute x=3
5into Equation 1 to solve for y.
33
5+ 2y−z= 4
9
5+ 2y−z= 4
2y−z= 1 −9
5
2y−z=−4
5
Step 4: Substitute x=3
5into Equation 3 to solve for z.
23
5+y+z= 6
6
5+y+z= 6
y+z= 6 −6
5
y+z=24
5
Step 5: Substitute y+z=24
5into 2y−z=−4
5to solve for y.
2y−24
5=−4
5
2y=24
5−4
5
2y=20
5
y= 2
Therefore, the solution to the system of equations is x=3
5,y= 2, and
z=14
5.
Question 27
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 5
25
Solution
Step 1: Let’s solve the second equation for y:
4x−y= 5 =⇒y= 4x−5
Step 2: Substitute the expression for yinto the first equation:
2x+ 3(4x−5) = 7
Step 3: Simplify and solve for x:
2x+ 12x−15 = 714x−15 = 714x= 22x=22
14 =11
7
Step 4: Substitute the value of xback into the equation y= 4x−5 to find
y:
y= 4 11
7−5 = 44
7−35
7=9
7
Therefore, the solution to the system of equations is x=11
7and y=9
7.
Question 28
Question
Solve the following system of equations:
(2x+ 3y= 7
3x−2y= 4
Solution
Step 1: We will use the method of substitution to solve the system of equations.
Step 2: Solve the first equation for x:
2x+ 3y= 7 =⇒x=7−3y
2
Step 3: Substitute this expression for xinto the second equation:
37−3y
2−2y= 4
Step 4: Simplify the equation:
21 −9y
2−2y= 4
Step 5: Multiply through by 2 to clear the fraction:
21 −9y−4y= 8
26
Step 6: Combine like terms:
21 −13y= 8
Step 7: Solve for y:
−13y=−13
Step 8: Divide by -13 to find y:
y= 1
Step 9: Substitute y= 1 back into the equation we found for x:
x=7−3(1)
2=4
2= 2
Step 10: Therefore, the solution to the system of equations is (x, y) = (2,1).
Question 29
Question
Solve the following system of equations:
(3x−2y= 7
2x+y= 4
Solution
Step 1: We will solve the second equation for y:
2x+y= 4
y= 4 −2x
Step 2: Substitute the expression for yinto the first equation:
3x−2(4 −2x) = 7
3x−8+4x= 7
7x−8 = 7
Step 3: Solve the equation for x:
7x−8 = 7
7x= 15
x=15
7
27
Step 4: Substitute the value of xback into y= 4 −2xto find y:
y= 4 −215
7
y= 4 −30
7
y=28
7−30
7
y=−2
7
Therefore, the solution to the system of equations is x=15
7and y=−2
7.
Question 30
Question
Solve the following system of equations:
(2x+ 3y= 10
3x+ 2y= 8
Solution
Step 1: Let’s multiply the first equation by 3 and the second equation by 2 to
create opposite coefficients for xwhen we add the equations together.
(6x+ 9y= 30
6x+ 4y= 16
Step 2: Subtract the second equation from the first equation to eliminate x.
5y= 14
Step 3: Solve for y.
y=14
5
Step 4: Substitute the value of yback into either of the original equations.
Let’s use the first equation.
2x+ 3 14
5= 10
Step 5: Solve for x.
2x+42
5= 10
2x= 10 −42
5
28
2x=8
5
x=4
5
Step 6: The solution to the system of equations is x=4
5and y=14
5.
Question 31
Question
Solve the following system of equations:
(3x+ 2y= 10
2x−3y=−5
Solution
Step 1: We will solve the system of equations using the method of elimination.
First, we will multiply the first equation by 3 and the second equation by 2 in
order to create opposite coefficients of y.
Step 2: Multiplying the first equation by 3, we get:
(9x+ 6y= 30
2x−3y=−5
Step 3: Multiplying the second equation by 2, we get:
(9x+ 6y= 30
4x−6y=−10
Step 4: Now we will add the two equations together to eliminate y:
(9x+ 6y) + (4x−6y) = 30 + (−10)
13x= 20
x=20
13
29
Step 5: Substituting x=20
13 back into the first equation, we can solve for y:
320
13+ 2y= 10
60
13 + 2y= 10
2y= 10 −60
13
2y=130
13 −60
13
2y=70
13
y=35
13
Step 6: Therefore, the solution to the system of equations is:
x=20
13, y =35
13
Question 32
Question
Solve the following system of equations:
(2x−3y= 5
4x+y= 3
Solution
Step 1: Solve the second equation for y:
4x+y= 3 =⇒y= 3 −4x
Step 2: Substitute y= 3 −4xinto the first equation and solve for x:
2x−3(3 −4x) = 5
2x−9 + 12x= 5
14x−9=5
14x= 14
x= 1
Step 3: Substitute x= 1 back into the second equation and solve for y:
4(1) + y= 3
30
4 + y= 3
y= 3 −4
y=−1
Step 4: Verify the solution by substituting x= 1 and y=−1 into both
original equations:
(2(1) −3(−1) = 5 (True)
4(1) + (−1) = 3 (True)
Therefore, the solution to the system of equations is x= 1 and y=−1.
Question 33
Question
Solve the following system of equations:
(2x−5y= 17
3x+ 2y= 1
Solution
Step 1: Let’s begin by multiplying the first equation by 2 and the second equa-
tion by 5 to eliminate y.
Multiplying the first equation by 2:
(4x−10y= 34
3x+ 2y= 1
Multiplying the second equation by 5:
(4x−10y= 34
15x+ 10y= 5
Step 2: Now, we will add the two new equations to eliminate y:
(4x−10y= 34
15x+ 10y= 5
(4x−10y) + (15x+ 10y) = 34 + 5
19x= 39
x=39
19
31
Step 3: Next, we will substitute the value of xback into one of the original
equations to solve for y. Let’s use the first equation 2x−5y= 17:
239
19−5y= 17
78
19 −5y= 17
−5y= 17 −78
19
−5y=323
19 −78
19
−5y=245
19
y=245
−19 ·5
y=−49
19
Step 4: Therefore, the solution to the system of equations is x=39
19 and
y=−49
19 .
Question 34
Question
Solve the following system of equations:
(2x+ 3y= 7
3x−2y= 9
Solution
Step 1: Let’s solve the system of equations by elimination. Multiply the first
equation by 2 and the second equation by 3 to make the coefficients of yequal:
(4x+ 6y= 14
9x−6y= 27
Step 2: Add the modified equations to eliminate y:
13x= 41
Step 3: Solve for x:
x=41
13 = 3.15
32
Step 4: Substitute xback into one of the original equations, for example,
2x+ 3y= 7, to solve for y:
2(3.15) + 3y= 7
6.3+3y= 7
3y= 0.7
y=0.7
3= 0.23
Step 5: The solution to the system of equations is x= 3.15 and y= 0.23.
Question 35
Question
Solve the following system of equations:
(2x+y= 5
3x−2y=−2
Solution
Step 1: We can solve this system of equations using the method of substitution.
Let’s solve the first equation for y:
y= 5 −2x
Step 2: Substitute y= 5 −2xinto the second equation and solve for x:
3x−2(5 −2x) = −2
Step 3: Simplify and solve for x:
3x−10 + 4x=−2
7x−10 = −2
7x= 8
x=8
7
Step 4: Substitute x=8
7back into y= 5 −2xto find y:
y= 5 −28
7
y= 5 −16
7
33
y=35
7−16
7
y=19
7
Step 5: Therefore, the solution to the system of equations is:
(x=8
7
y=19
7
34
