MATH 114 - QUANTITATIVE
REASONING - Systems of Equations
Question Bank - Set 3
Liberty University
Question 1
Question
Solve the following system of equations for xand y:
(4x−3y= 11
2x+ 5y=−7
Solution
Step 1: Start by solving the first equation for x. Step 2: Solve for xin terms of
yby rearranging the equation. Step 3: Substitute the expression for xinto the
second equation to solve for y. Step 4: Substitute the value of yback into the
equation solved for xto find its value. Step 5: Verify the solution by checking
if it satisfies both original equations.
Step 1: Solve the first equation for x:
4x−3y= 11 =⇒x=3y+ 11
4
Step 2: Solve for xin terms of yby rearranging the equation:
x=3y+ 11
4
Step 3: Substitute the expression for xinto the second equation to solve
for y:
23y+ 11
4+ 5y=−7
Solving for y, we get: 3
2y+11
2+ 5y=−7
13
2y=−18
y=−36
13
Step 4: Substitute the value of yback into the equation solved for xto find
its value:
x=3(−36
13 ) + 11
4=−5
13
Step 5: Verify the solution by checking if it satisfies both original equations:
(4(−5
13 )−3(−36
13 ) = 11
2(−5
13 ) + 5(−36
13 ) = −7
Both equations are satisfied, therefore the solution to the system of equations
is x=−5
13 and y=−36
13 .
Question 2
Question
Solve the following system of equations:
(3x−2y= 7
2x+ 5y= 3
Solution
Step 1: Let’s first solve the first equation for x:
3x−2y= 7
3x= 2y+ 7
x=2y+ 7
3
Step 2: Substitute the expression for xinto the second equation:
22y+ 7
3+ 5y= 3
4y+ 14
3+ 5y= 3
4y+ 14 + 15y= 9
19y+ 14 = 9
19y=−5
2
y=−5
19
Step 3: Substitute yback into the expression we found for x:
x=2(−5
19 )+7
3
x=−10/19 + 7
3
x=−10/19 + 7 ∗19/19
3
x=−10 + 133
57
x=123
57 =41
19
Step 4: Therefore, the solution to the system of equations is:
x=41
19, y =−5
19
Question 3
Question
Solve the following system of equations:
(3x−2y= 7
5x+ 4y= 19
Solution
Step 1: Multiply the first equation by 2 and the second equation by 3 to
eliminate ywhen adding the equations.
(6x−4y= 14
15x+ 12y= 57
Step 2: Add the two equations to obtain a new equation in terms of x.
21x= 71
Step 3: Solve for x.
x=71
21 =71
21 =71
21 =71
21
3
Step 4: Substitute the value of xback into one of the original equations to
solve for y. Using the first equation:
371
21−2y= 7
Step 5: Solve for y.
y=5
3
Step 6: Therefore, the solution to the system of equations is x=71
21 and
y=5
3.
Question 4
Question
Solve the following system of equations:
3x−2y+z= 5
2x+ 3y−z= 3
x+ 4y+ 2z= 7
Solution
Step 1: Start by choosing two equations to eliminate a variable. Step 2: Multiply
the first equation by 2 and the second equation by 3 to eliminate z. Step 3:
Add the resulting equations to eliminate zand solve for x. Step 4: Substitute
the value of xback into one of the original equations to solve for y. Step 5:
Substitute the values of xand yback into one of the original equations to solve
for z.
Step 1: We will choose to eliminate zfirst.
Step 2: Multiply the first equation by 2 and the second equation by 3:
(6x−4y+ 2z= 10
6x+ 9y−3z= 9
Step 3: Add the resulting equations to eliminate z:
15y= 19 ⇒y=19
15
Step 4: Substitute yback into the first equation to solve for x:
3x−219
15+z= 5 ⇒x=16
15
4
Step 5: Substitute xand yback into the third equation to solve for z:
16
15 + 4 19
15+ 2z= 7 ⇒z=11
15
Therefore, the solution to the system of equations is x=16
15 ,y=19
15 , and
z=11
15 .
Question 5
Question
Solve the system of equations:
(2x+ 3y= 11
4x−5y=−7
Solution
Step 1: Start by multiplying the first equation by 2 to eliminate xwhen adding
the two equations together.
(4x+ 6y= 22
4x−5y=−7
Step 2: Subtract the second equation from the first equation.
11y= 29
Step 3: Solve for yby dividing by 11.
y=29
11 =29
11
Step 4: Substitute yback into the first equation to solve for x.
2x+ 3 29
11= 11
Step 5: Solve for x.
2x+87
11 = 11
2x=11 ·11 −87
11
2x=121 −87
11 =34
11
x=34
22 =17
11
Therefore, the solution to the system of equations is x=17
11 and y=29
11 .
5
Question 6
Question
Solve the following system of equations:
3x+ 2y−z= 7
x−4y+ 2z=−1
2x+y+ 3z= 9
Solution
Step 1: Let’s start by writing the system of equations in matrix form:
3 2 −1
1−4 2
2 1 3
x
y
z
=
7
−1
9
Step 2: We will now perform row operations to simplify the matrix.
3 2 −1|7
1−4 2 | −1
2 1 3 |9
Step 3: Subtract 2 times the second row from the third row to eliminate the
xterm:
3 2 −1|7
1−4 2 | −1
0 9 −1|11
Step 4: Divide the third row by 9 to simplify calculations:
3 2 −1|7
1−4 2 | −1
0 1 −1
9|11
9
Step 5: Add 4 times the third row to the second row to eliminate the yterm:
3 2 −1|7
1 0 7
9|39
9
0 1 −1
9|11
9
Step 6: Subtract 2 times the second row from the first row to eliminate the
xterm:
1 2 −5
9| −5
1 0 7
9|39
9
0 1 −1
9|11
9
Step 7: Subtract the first row from the second row to get the value of x:
x=−2
6
Step 8: Substitute x=−2 into the second row to solve for y:
y=5
3
Step 9: Substitute x=−2 and y=5
3into the first row to solve for z:
z= 3
Therefore, the solution to the system of equations is x=−2, y=5
3, and
z= 3.
Question 7
Question
Solve the following system of equations:
(2x+ 3y= 11
4x−y= 4
Solution
Step 1: Solve the second equation for y:
4x−y= 4 =⇒y= 4x−4
Step 2: Substitute the expression for yinto the first equation:
2x+ 3(4x−4) = 11
Step 3: Simplify the equation:
2x+ 12x−12 = 11
Step 4: Combine like terms:
14x−12 = 11
Step 5: Add 12 to both sides:
14x= 23
Step 6: Divide by 14:
x=23
14
Step 7: Substitute the value of xback into y= 4x−4:
y= 4 23
14−4
7
Step 8: Simplify the expression:
y=46
7−4 = 46 −28
7=18
7
Step 9: Therefore, the solution to the system of equations is x=23
14 and
y=18
7.
Question 8
Question
Solve the following system of equations:
(2x+ 3y= 5
4x−y= 7
Solution
Step 1: Let’s first solve the second equation for yin terms of x.
4x−y= 7 =⇒y= 4x−7
Step 2: Now, substitute y= 4x−7 into the first equation and solve for x.
2x+ 3(4x−7) = 5
Step 3: Simplify and solve for x.
2x+ 12x−21 = 5 =⇒14x= 26 =⇒x=13
7
Step 4: Now that we have found x, substitute it back into y= 4x−7 to
find y.
y= 4 13
7−7 = 52
7−7 = 52 −49
7=3
7
Step 5: Therefore, the solution to the system of equations is x=13
7and
y=3
7.
Question 9
Question
Solve the following system of equations:
(3x+ 2y= 7
4x−3y= 1
8
Solution
Step 1: Multiply the first equation by 3 and the second equation by 2 to elimi-
nate y.
(9x+ 6y= 21
8x−6y= 2
Step 2: Add the two modified equations.
17x= 23
Step 3: Solve for x.
x=23
17
Step 4: Substitute xback into one of the original equations, for example,
the first one, to solve for y.
323
17+ 2y= 7
Step 5: Simplify and solve for y.
y=5
17
Therefore, the solution to the system of equations is x=23
17 and y=5
17 .
Question 10
Question
Find the solution to the following system of linear equations:
(2x−3y= 7
4x+ 5y= 1
Solution
Step 1: Start by multiplying the first equation by 2 to eliminate xwhen adding
the equations.
(4x−6y= 14
4x+ 5y= 1
Step 2: Subtract the equations to eliminate x.
−11y=−13
Step 3: Solve for y.
y=−13
−11 =13
11
9
Step 4: Substitute yback into one of the original equations and solve for x.
2x−313
11= 7
2x−39
11 = 7
2x=116
11
x=58
11
Step 5: Therefore, the solution to the system of equations is x=58
11 and
y=13
11 .
Question 11
Question
Solve the following system of equations:
(2x+ 3y= 5
4x−2y= 10
Solution
Step 1: Begin by multiplying the first equation by 2 to eliminate xwhen adding
the two equations together:
(4x+ 6y= 10
4x−2y= 10
Step 2: Subtract the second equation from the first equation:
8y= 0
Step 3: Solve for y:
y= 0
Step 4: Substitute y= 0 back into the first equation to solve for x:
2x+ 3(0) = 5 =⇒2x= 5 =⇒x=5
2
Step 5: The solution to the system of equations is x=5
2and y= 0.
10
Question 12
Question
Solve the following system of equations:
(3x+ 2y= 10
5x−4y= 8
Solution
Step 1: Start by solving the first equation for xin terms of y.
3x+ 2y= 10
3x= 10 −2y
x=10 −2y
3
Step 2: Substitute the expression found for xinto the second equation.
510 −2y
3−4y= 8
50 −10y
3−4y= 8
50 −10y−12y
3= 8
50 −22y= 24
−22y=−26
y=26
22
y=13
11
Step 3: Substitute the value of yback into the expression found for xto find
the value of x.
x=10 −213
11
3
x=10 −26
11
3
x=110 −26
33
x=84
33
x=28
11
Therefore, the solution to the system of equations is x=28
11 and y=13
11 .
11
Question 13
Question
Solve the following system of equations using the method of your choice:
(2x+ 3y= 11
4x−y= 5
Solution
To solve this system of equations, we can use the substitution method.
Step 1: Solve the second equation for y:
4x−y= 5 =⇒y= 4x−5
Step 2: Substitute y= 4x−5 into the first equation:
2x+ 3(4x−5) = 11
Step 3: Simplify the equation:
2x+ 12x−15 = 11 =⇒14x−15 = 11
Step 4: Solve for x:
14x= 26 =⇒x=26
14 =13
7
Step 5: Substitute x=13
7back into the equation y= 4x−5 to solve for y:
y= 4 13
7−5 = 52
7−5 = 52 −35
7=17
7
Step 6: The solution to the system of equations is:
(x=13
7
y=17
7
Question 14
Question
Solve the following system of equations:
(3x+ 4y= 10
2x−y= 3
12
Solution
Step 1: Solve the second equation for yin terms of x.
2x−y= 3
y= 2x−3
Step 2: Substitute y= 2x−3 into the first equation and solve for x.
3x+ 4(2x−3) = 10
3x+ 8x−12 = 10
11x= 22
x= 2
Step 3: Substitute x= 2 back into the second equation and solve for y.
2(2) −y= 3
4−y= 3
y= 1
Therefore, the solution to the system of equations is x= 2 and y= 1.
Question 15
Question
Solve the following system of equations:
(2x−3y= 7
3x+ 2y= 5
Solution
Step 1: Multiply the first equation by 2 and the second equation by 3 to elimi-
nate y.
(4x−6y= 14
9x+ 6y= 15
Step 2: Add the equations together to eliminate y.
13x= 29
Step 3: Solve for x.
x=29
13
13
Step 4: Substitute xback into one of the original equations to solve for y.
Let’s substitute into the first equation.
229
13−3y= 7
Step 5: Simplify and solve for y.
58
13 −3y= 7
58
13 −39
13 = 3y
19
13 = 3y
y=19
39
Step 6: The solution to the system of equations is x=29
13 and y=19
39 .
Question 16
Question
Solve the following system of equations:
(3x−2y= 7
2x+y=−5
Solution
Step 1: Solve the second equation for yin terms of x.
2x+y=−5
y=−2x−5
Step 2: Substitute y=−2x−5 into the first equation and solve for x.
3x−2(−2x−5) = 7
3x+ 4x+ 10 = 7
7x=−3
x=−3
7
14
Step 3: Substitute x=−3
7back into y=−2x−5 to find the value of y.
y=−2−3
7−5
y=6
7−5
y=−29
7
Therefore, the solution to the system of equations is x=−3
7and y=−29
7.
Question 17
Question
Solve the following system of equations:
(x2+y2= 29
x−y= 1
Solution
Step 1: Solve the second equation for xin terms of y.
x−y= 1
x=y+ 1
Step 2: Substitute xin terms of yinto the first equation.
(y+ 1)2+y2= 29
y2+ 2y+1+y2= 29
2y2+ 2y−28 = 0
y2+y−14 = 0
Step 3: Solve the quadratic equation y2+y−14 = 0.
(y+ 7)(y−2) = 0
y=−7,2
Step 4: Find the corresponding values of xusing x=y+ 1.
When y=−7, x =−6
When y= 2, x = 3
Thus, the solutions to the system of equations are (3,2) and (−6,−7).
15
Question 18
Question
Solve the following system of equations:
(3x+ 2y= 4
5x−3y=−7
Solution
Step 1: Multiply the first equation by 3 and the second equation by 2 to elimi-
nate y.
(9x+ 6y= 12
10x−6y=−14
Step 2: Add the two equations together to eliminate y.
19x=−2
Step 3: Solve for x.
x=−2
19
Step 4: Substitute xback into one of the original equations to solve for y.
Let’s use the first equation.
3−2
19 + 2y= 4
Step 5: Simplify and solve for y.
−6/19 + 2y= 4
2y= 4 + 6/19
y=38
19
So, the solution to the system of equations is x=−2
19 and y=38
19 .
Question 19
Question
Solve the following system of equations:
(3x+ 2y= 8
2x−4y=−5
16
Solution
Step 1: We will use the method of substitution to solve this system of equations.
Step 2: Let’s solve the first equation for x:
3x+ 2y= 8 =⇒x=8−2y
3
Step 3: Substitute xfrom the first equation into the second equation:
28−2y
3−4y=−5
Step 4: Simplify the equation:
16 −4y
3−4y=−5
16 −4y−12y=−15
16 −16y=−15
−16y=−31
y=31
16
Step 5: Substitute yback into the equation x=8−2y
3to find x:
x=8−231
16
3
x=8−62
16
3
x=8−62
16
3
x=8−31
8
3
x=64 −31
24
x=33
24
x=11
8
Step 6: Therefore, the solution to the system of equations is x=11
8and y=31
16 .
17
Question 20
Question
Solve the following system of equations:
(3x−2y= 7
6x−4y= 14
Solution
Step 1: We can start by multiplying the first equation by 2 to make the coeffi-
cients of ythe same in both equations:
(6x−4y= 14
6x−4y= 14
Step 2: Subtracting the second equation from the first, we get:
0=0
Step 3: Since we obtained a trivial equation, this means that the system of
equations have infinite solutions and the equations are dependent. This implies
that the two equations represent the same line in the plane.
Therefore, the system of equations has infinitely many solutions, and the
solution set can be written as follows:
{(x, y)|3x−2y= 7}
Question 21
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 6
Solution
Step 1: Solve the second equation for y:
4x−y= 6
y= 4x−6
18
Step 2: Substitute yin the first equation:
2x+ 3(4x−6) = 7
2x+ 12x−18 = 7
14x= 25
x=25
14
Step 3: Substitute xback into y= 4x−6 to find y:
y= 4 25
14−6
y=100
14 −84
14
y=16
14
y=8
7
Step 4: The solution to the system of equations is x=25
14 and y=8
7.
Question 22
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
14x= 22
x=22
14
x=11
7
19
Step 3: Substitute x=11
7back into y= 4x−5 to find y.
y= 4 11
7−5
y=44
7−5
y=44
7−35
7
y=9
7
Therefore, the solution to the system of equations is x=11
7and y=9
7.
Question 23
Question
Find the solution to the system of equations:
(3x+ 2y= 5
x2+xy +y2= 7
Solution
To solve this system of equations, we will first use the first equation to express
xin terms of y, then substitute it into the second equation.
Step 1: Solve the first equation for x.
3x+ 2y= 5
3x= 5 −2y
x=5−2y
3
Step 2: Substitute xinto the second equation.
5−2y
32
+5−2y
3y+y2= 7
(5 −2y)2
9+(5 −2y)y
3+y2= 7
25 −20y+ 4y2
9+5y−2y2
3+y2= 7
25 −20y+ 4y2+ 15y−6y2+ 9y2
9= 7
25 −5y+ 7y2
9= 7
25 −5y+ 7y2= 63
7y2−5y−38 = 0
20
Step 3: Solve the quadratic equation. Using the quadratic formula
y=−b±√b2−4ac
2a, we find:
y=5±p(−5)2−4(7)(−38)
2(7)
y=5±√25 + 1064
14
y=5±√1089
14
y=5±33
14
So, y=38
14 =19
7or y=−28
14 =−2.
Step 4: Find the corresponding xvalues for each yvalue. Using
x=5−2y
3: For y=19
7:
x=5−219
7
3
x=5−38
7
3
x=35 −38
21
x=−3
21
x=−1
7
For y=−2:
x=5−2(−2)
3
x=5+4
3
x=9
3
x= 3
Thus, the solutions to the system of equations are x=−1
7,y=19
7and
x= 3, y=−2.
21
Question 24
Question
Consider the following system of equations:
2x+ 3y+z= 4
x−y+ 2z= 3
3x+ 2y−z= 2
Determine whether the system has a unique solution, infinitely many solu-
tions, or no solution. If a solution exists, find the solution set.
Solution
Step 1: First, let’s write the system of equations in matrix form:
2 3 1
1−1 2
3 2 −1
x
y
z
=
4
3
2
Step 2: Next, we will find the determinant of the coefficient matrix on the
left side:
det
2 3 1
1−1 2
3 2 −1
=−22
Since the determinant is not zero, the system has a unique solution.
Step 3: To find the solution set, we will use Cramer’s Rule. Calculate the
determinants of the matrices formed by replacing the columns of the coefficient
matrix one at a time with the column vector on the right side:
Dx= det
4 3 1
3−1 2
2 2 −1
= 39
Dy= det
2 4 1
1 3 2
3 2 −1
=−28
Dz= det
234
1−1 3
322
= 18
Step 4: Finally, we can find the values of x,y, and zusing the formulas
x=Dx
det ,y=Dy
det , and z=Dz
det :
x=39
−22 =−39
22
22
y=−28
−22 =14
11
z=18
−22 =−9
11
Therefore, the unique solution to the system of equations is (x, y, z) =
−39
22 ,14
11 ,−9
11 .
Question 25
Question
Solve the following system of equations:
(2x−3y= 1
3x+ 4y= 4
Solution
Step 1: Begin by solving the first equation for xin terms of y.
2x−3y= 1 =⇒2x= 3y+ 1 =⇒x=3y+ 1
2
Step 2: Substitute xin the second equation with the expression 3y+1
2.
33y+ 1
2+ 4y= 4
Step 3: Simplify the expression and solve for y.
9y+ 3
2+4y= 4 =⇒9y+3+8y= 8 =⇒17y+3 = 8 =⇒17y= 5 =⇒y=5
17
Step 4: Substitute yback into the equation to solve for x.
x=3( 5
17 )+1
2=15
17 +17
17 =32
17
Step 5: Therefore, the solution to the system of equations is:
(x=32
17
y=5
17
Question 26
Question
Solve the following system of equations:
(2x−3y= 5
4x+ky = 9
23
Solution
Step 1: Let’s start by multiplying the first equation by 2 to eliminate xwhen
adding the equations.
(4x−6y= 10
4x+ky = 9
Step 2: Subtract the modified first equation from the second equation to
solve for y.
(4x+ky)−(4x−6y)=9−10
(4x−4x)+(ky + 6y) = −1
ky + 6y=−1
y(k+ 6) = −1
y=−1
k+ 6
Step 3: Substitute the expression for yback into the first equation and solve
for x.
2x−3−1
k+ 6= 5
2x+3
k+ 6 = 5
2x= 5 −3
k+ 6
2x=5(k+ 6) −3
k+ 6
2x=5k+ 30 −3
k+ 6
2x=5k+ 27
k+ 6
x=5k+ 27
2(k+ 6)
Therefore, the solution to the system of equations is:
(x=5k+27
2(k+6)
y=−1
k+6
Question 27
Question
Solve the following system of equations:
(2x+ 3y= 7
3x−2y= 8
24
Solution
Step 1: Let’s first multiply the first equation by 3 and the second equation by
2 to make the coefficients of ymatch for elimination.
(6x+ 9y= 21
6x−4y= 16
Step 2: Now, subtract the second equation from the first equation to elimi-
nate x:
13y= 5
Step 3: Solve for y:
y=5
13
Step 4: Substitute y=5
13 back into one of the original equations (let’s use
the first one) and solve for x:
2x+ 3 5
13= 7
2x+15
13 = 7
Step 5: Subtract 15
13 from both sides and solve for x:
2x=84
13
x=84
26
x=42
13
Step 6: The solution to the system of equations is x=42
13 and y=5
13 .
Question 28
Question
Solve the following system of equations:
(3x−2y= 5
2x+ 3y= 4
25
Solution
Step 1: Let’s use the elimination method to solve this system of equations. Our
goal is to eliminate one variable by adding or subtracting the equations.
Step 2: To eliminate y, multiply the first equation by 3 and the second
equation by 2:
(9x−6y= 15
4x+ 6y= 8
Step 3: Add the two equations:
13x= 23
Step 4: Solve for x:
x=23
13
Step 5: Substitute xback into the first equation to solve for y:
323
13−2y= 5
y=−4
13
Step 6: Therefore, the solution to the system of equations is:
(x=23
13
y=−4
13
Question 29
Question
Solve the following system of equations:
(3x+ 2y= 5
6x+ 4y= 10
Solution
Step 1: Multiply the first equation by 2 to create a system of equations with
the same coefficient for y.
(6x+ 4y= 10
6x+ 4y= 10
Step 2: Subtract one equation from the other to eliminate y.
(6x+ 4y)−(6x+ 4y) = 10 −10
0=0
26
Step 3: Since 0 = 0 is always true, the system has infinitely many solutions.
Thus, the solution set is all pairs of the form (x, 5−3x
2), where xis a real
number.
Question 30
Question
Solve the following system of equations:
(2x+ 3y= 7
3x+ 4y= 11
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+ 8y= 22
Step 2: Subtract the second equation from the first equation:
(6x+ 9y)−(6x+ 8y) = 21 −22
x+y=−1
Step 3: Substitute y=−1−xinto the first equation and solve for x:
2x+ 3(−1−x)=7
2x−3−3x= 7
−x−3=7
−x= 10
x=−10
Step 4: Substitute x=−10 into y=−1−xand solve for y:
y=−1−(−10)
y=−1 + 10
y= 9
Therefore, the solution to the system of equations is x=−10 and y= 9.
27
Question 31
Question
Solve the following system of equations:
3x+ 2y−z= 3
2x−y+z= 2
x+ 3y+ 2z= 4
Solution
Step 1: Let’s call the given system of equations (1), (2), and (3) respectively.
To solve the system, we will use the method of substitution.
Step 2: From equation (2), we can express zin terms of xand yas:
z= 2 −2x+y
Step 3: Substituting zin equations (1) and (3) with the expression 2−2x+y,
we get:
(3x+ 2y−(2 −2x+y) = 3
x+ 3y+ 2(2 −2x+y) = 4
Step 4: Simplifying equation (1) gives us:
5x+ 3y= 5
Step 5: Substituting equation (2) back into equation (3) gives us:
x+ 3y+ 4 −4x+ 2y= 4
Step 6: Simplifying equation (3) gives us:
−3x+ 5y= 0
Step 7: We now have a system of two equations:
(5x+ 3y= 5
−3x+ 5y= 0
Step 8: Solving the system of equations (4) and (5) simultaneously gives us
x= 5 and y=−5.
Step 9: Substituting x= 5 and y=−5 back into equation (2) to solve for z
gives us z= 2.
Step 10: Therefore, the solution to the system of equations is:
x= 5
y=−5
z= 2
28
Question 32
Question
Solve the following system of equations:
3x−2y+z= 4
2x+y−3z=−2
x−4y+ 5z= 8
Solution
Step 1: Multiply the second equation by 2 and the third equation by 3 to
eliminate xwhen adding/subtracting equations.
3x−2y+z= 4
4x+ 2y−6z=−4
3x−12y+ 15z= 24
Step 2: Subtract the first equation from the second equation to eliminate y.
3x−2y+z= 4
x−4z=−8
3x−12y+ 15z= 24
Step 3: Multiply the second equation by 3 and subtract it from the first
equation to eliminate x.
3x−2y+z= 4
x−4z=−8
3z= 36
Step 4: Solve the third equation to find z.
z= 12
Step 5: Substitute z= 12 back into the second equation to solve for x.
x−4(12) = −8
x−48 = −8
x= 40
Step 6: Substitute x= 40 and z= 12 back into the first equation to solve
for y.
3(40) −2y+ 12 = 4
29
120 −2y+ 12 = 4
−2y=−128
y= 64
Therefore, the solution to the system of equations is:
x= 40, y = 64, z = 12
Question 33
Question
Solve the following system of equations:
(2x+ 3y= 11
3x−2y= 1
Solution
Step 1: Multiply the first equation by 3 and the second equation by 2 to get:
(6x+ 9y= 33
6x−4y= 2
Step 2: Subtract the second equation from the first equation to eliminate x:
13y= 31
Step 3: Solve for y:
y=31
13 = 2.38
Step 4: Substitute yback into one of the original equations (let’s use the
first one):
2x+ 3(2.38) = 11
Step 5: Solve for x:
2x+ 7.14 = 11
Step 6: Solve for x:
2x= 3.86
Step 7: Solve for x:
x= 1.93
Therefore, the solution to the system of equations is x= 1.93 and y= 2.38.
30
Question 34
Question
Solve the following system of equations:
(2x+ 3y= 7
3x−4y=−4
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 the expression for xinto the second equation:
37−3y
2−4y=−4
Step 3: Solve the equation for y:
21 −9y
2−4y=−4
21 −9y−8y=−8
21 −17y=−8
−17y=−29
y=29
17
Step 4: Substitute the value of yback into the equation to solve for x:
x=7−329
17
2
x=7−87
17
2
x=7·17 −87
34
x=119 −87
34
x=32
34
x=16
17
Therefore, the solution to the system of equations is x=16
17 and y=29
17 .
31
Question 35
Question
Solve the following system of equations:
3x−y+ 2z= 5
x+ 4y−z=−3
2x−3y+ 2z= 6
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 yin terms of xand z.
3x−y+ 2z= 5
y= 3x+ 2z−5
Step 2: Next, substitute y= 3x+ 2z−5 into the second equation and solve
for x.
x+ 4(3x+ 2z−5) −z=−3
x+ 12x+ 8z−20 −z=−3
13x+ 7z= 17
13x=−7z+ 17
x=−7
13z+17
13
Step 3: Now substitute x=−7
13 z+17
13 back into y= 3x+ 2z−5 to find y
in terms of z.
y= 3 −7
13z+17
13+ 2z−5
y=−21
13z+51
13 + 2z−5
y=−21
13z+ 2z+51
13 −65
13
y=−7
13z−14
13
Step 4: Substitute x=−7
13 z+17
13 and y=−7
13 z−14
13 into the third equation
32
13
2y=−18
y=−36
13
Step 4: Substitute the value of yback into the equation solved for xto find
its value:
x=3(−36
13 ) + 11
4=−5
13
Step 5: Verify the solution by checking if it satisfies both original equations:
(4(−5
13 )−3(−36
13 ) = 11
2(−5
13 ) + 5(−36
13 ) = −7
Both equations are satisfied, therefore the solution to the system of equations
is x=−5
13 and y=−36
13 .
Question 2
Question
Solve the following system of equations:
(3x−2y= 7
2x+ 5y= 3
Solution
Step 1: Let’s first solve the first equation for x:
3x−2y= 7
3x= 2y+ 7
x=2y+ 7
3
Step 2: Substitute the expression for xinto the second equation:
22y+ 7
3+ 5y= 3
4y+ 14
3+ 5y= 3
4y+ 14 + 15y= 9
19y+ 14 = 9
19y=−5
2
y=−5
19
Step 3: Substitute yback into the expression we found for x:
x=2(−5
19 )+7
3
x=−10/19 + 7
3
x=−10/19 + 7 ∗19/19
3
x=−10 + 133
57
x=123
57 =41
19
Step 4: Therefore, the solution to the system of equations is:
x=41
19, y =−5
19
Question 3
Question
Solve the following system of equations:
(3x−2y= 7
5x+ 4y= 19
Solution
Step 1: Multiply the first equation by 2 and the second equation by 3 to
eliminate ywhen adding the equations.
(6x−4y= 14
15x+ 12y= 57
Step 2: Add the two equations to obtain a new equation in terms of x.
21x= 71
Step 3: Solve for x.
x=71
21 =71
21 =71
21 =71
21
3
Step 4: Substitute the value of xback into one of the original equations to
solve for y. Using the first equation:
371
21−2y= 7
Step 5: Solve for y.
y=5
3
Step 6: Therefore, the solution to the system of equations is x=71
21 and
y=5
3.
Question 4
Question
Solve the following system of equations:
3x−2y+z= 5
2x+ 3y−z= 3
x+ 4y+ 2z= 7
Solution
Step 1: Start by choosing two equations to eliminate a variable. Step 2: Multiply
the first equation by 2 and the second equation by 3 to eliminate z. Step 3:
Add the resulting equations to eliminate zand solve for x. Step 4: Substitute
the value of xback into one of the original equations to solve for y. Step 5:
Substitute the values of xand yback into one of the original equations to solve
for z.
Step 1: We will choose to eliminate zfirst.
Step 2: Multiply the first equation by 2 and the second equation by 3:
(6x−4y+ 2z= 10
6x+ 9y−3z= 9
Step 3: Add the resulting equations to eliminate z:
15y= 19 ⇒y=19
15
Step 4: Substitute yback into the first equation to solve for x:
3x−219
15+z= 5 ⇒x=16
15
4
Step 5: Substitute xand yback into the third equation to solve for z:
16
15 + 4 19
15+ 2z= 7 ⇒z=11
15
Therefore, the solution to the system of equations is x=16
15 ,y=19
15 , and
z=11
15 .
Question 5
Question
Solve the system of equations:
(2x+ 3y= 11
4x−5y=−7
Solution
Step 1: Start by multiplying the first equation by 2 to eliminate xwhen adding
the two equations together.
(4x+ 6y= 22
4x−5y=−7
Step 2: Subtract the second equation from the first equation.
11y= 29
Step 3: Solve for yby dividing by 11.
y=29
11 =29
11
Step 4: Substitute yback into the first equation to solve for x.
2x+ 3 29
11= 11
Step 5: Solve for x.
2x+87
11 = 11
2x=11 ·11 −87
11
2x=121 −87
11 =34
11
x=34
22 =17
11
Therefore, the solution to the system of equations is x=17
11 and y=29
11 .
5
Question 6
Question
Solve the following system of equations:
3x+ 2y−z= 7
x−4y+ 2z=−1
2x+y+ 3z= 9
Solution
Step 1: Let’s start by writing the system of equations in matrix form:
3 2 −1
1−4 2
2 1 3
x
y
z
=
7
−1
9
Step 2: We will now perform row operations to simplify the matrix.
3 2 −1|7
1−4 2 | −1
2 1 3 |9
Step 3: Subtract 2 times the second row from the third row to eliminate the
xterm:
3 2 −1|7
1−4 2 | −1
0 9 −1|11
Step 4: Divide the third row by 9 to simplify calculations:
3 2 −1|7
1−4 2 | −1
0 1 −1
9|11
9
Step 5: Add 4 times the third row to the second row to eliminate the yterm:
3 2 −1|7
1 0 7
9|39
9
0 1 −1
9|11
9
Step 6: Subtract 2 times the second row from the first row to eliminate the
xterm:
1 2 −5
9| −5
1 0 7
9|39
9
0 1 −1
9|11
9
Step 7: Subtract the first row from the second row to get the value of x:
x=−2
6
Step 8: Substitute x=−2 into the second row to solve for y:
y=5
3
Step 9: Substitute x=−2 and y=5
3into the first row to solve for z:
z= 3
Therefore, the solution to the system of equations is x=−2, y=5
3, and
z= 3.
Question 7
Question
Solve the following system of equations:
(2x+ 3y= 11
4x−y= 4
Solution
Step 1: Solve the second equation for y:
4x−y= 4 =⇒y= 4x−4
Step 2: Substitute the expression for yinto the first equation:
2x+ 3(4x−4) = 11
Step 3: Simplify the equation:
2x+ 12x−12 = 11
Step 4: Combine like terms:
14x−12 = 11
Step 5: Add 12 to both sides:
14x= 23
Step 6: Divide by 14:
x=23
14
Step 7: Substitute the value of xback into y= 4x−4:
y= 4 23
14−4
7
Step 8: Simplify the expression:
y=46
7−4 = 46 −28
7=18
7
Step 9: Therefore, the solution to the system of equations is x=23
14 and
y=18
7.
Question 8
Question
Solve the following system of equations:
(2x+ 3y= 5
4x−y= 7
Solution
Step 1: Let’s first solve the second equation for yin terms of x.
4x−y= 7 =⇒y= 4x−7
Step 2: Now, substitute y= 4x−7 into the first equation and solve for x.
2x+ 3(4x−7) = 5
Step 3: Simplify and solve for x.
2x+ 12x−21 = 5 =⇒14x= 26 =⇒x=13
7
Step 4: Now that we have found x, substitute it back into y= 4x−7 to
find y.
y= 4 13
7−7 = 52
7−7 = 52 −49
7=3
7
Step 5: Therefore, the solution to the system of equations is x=13
7and
y=3
7.
Question 9
Question
Solve the following system of equations:
(3x+ 2y= 7
4x−3y= 1
8
Solution
Step 1: Multiply the first equation by 3 and the second equation by 2 to elimi-
nate y.
(9x+ 6y= 21
8x−6y= 2
Step 2: Add the two modified equations.
17x= 23
Step 3: Solve for x.
x=23
17
Step 4: Substitute xback into one of the original equations, for example,
the first one, to solve for y.
323
17+ 2y= 7
Step 5: Simplify and solve for y.
y=5
17
Therefore, the solution to the system of equations is x=23
17 and y=5
17 .
Question 10
Question
Find the solution to the following system of linear equations:
(2x−3y= 7
4x+ 5y= 1
Solution
Step 1: Start by multiplying the first equation by 2 to eliminate xwhen adding
the equations.
(4x−6y= 14
4x+ 5y= 1
Step 2: Subtract the equations to eliminate x.
−11y=−13
Step 3: Solve for y.
y=−13
−11 =13
11
9
Step 4: Substitute yback into one of the original equations and solve for x.
2x−313
11= 7
2x−39
11 = 7
2x=116
11
x=58
11
Step 5: Therefore, the solution to the system of equations is x=58
11 and
y=13
11 .
Question 11
Question
Solve the following system of equations:
(2x+ 3y= 5
4x−2y= 10
Solution
Step 1: Begin by multiplying the first equation by 2 to eliminate xwhen adding
the two equations together:
(4x+ 6y= 10
4x−2y= 10
Step 2: Subtract the second equation from the first equation:
8y= 0
Step 3: Solve for y:
y= 0
Step 4: Substitute y= 0 back into the first equation to solve for x:
2x+ 3(0) = 5 =⇒2x= 5 =⇒x=5
2
Step 5: The solution to the system of equations is x=5
2and y= 0.
10
Question 12
Question
Solve the following system of equations:
(3x+ 2y= 10
5x−4y= 8
Solution
Step 1: Start by solving the first equation for xin terms of y.
3x+ 2y= 10
3x= 10 −2y
x=10 −2y
3
Step 2: Substitute the expression found for xinto the second equation.
510 −2y
3−4y= 8
50 −10y
3−4y= 8
50 −10y−12y
3= 8
50 −22y= 24
−22y=−26
y=26
22
y=13
11
Step 3: Substitute the value of yback into the expression found for xto find
the value of x.
x=10 −213
11
3
x=10 −26
11
3
x=110 −26
33
x=84
33
x=28
11
Therefore, the solution to the system of equations is x=28
11 and y=13
11 .
11
Question 13
Question
Solve the following system of equations using the method of your choice:
(2x+ 3y= 11
4x−y= 5
Solution
To solve this system of equations, we can use the substitution method.
Step 1: Solve the second equation for y:
4x−y= 5 =⇒y= 4x−5
Step 2: Substitute y= 4x−5 into the first equation:
2x+ 3(4x−5) = 11
Step 3: Simplify the equation:
2x+ 12x−15 = 11 =⇒14x−15 = 11
Step 4: Solve for x:
14x= 26 =⇒x=26
14 =13
7
Step 5: Substitute x=13
7back into the equation y= 4x−5 to solve for y:
y= 4 13
7−5 = 52
7−5 = 52 −35
7=17
7
Step 6: The solution to the system of equations is:
(x=13
7
y=17
7
Question 14
Question
Solve the following system of equations:
(3x+ 4y= 10
2x−y= 3
12
Solution
Step 1: Solve the second equation for yin terms of x.
2x−y= 3
y= 2x−3
Step 2: Substitute y= 2x−3 into the first equation and solve for x.
3x+ 4(2x−3) = 10
3x+ 8x−12 = 10
11x= 22
x= 2
Step 3: Substitute x= 2 back into the second equation and solve for y.
2(2) −y= 3
4−y= 3
y= 1
Therefore, the solution to the system of equations is x= 2 and y= 1.
Question 15
Question
Solve the following system of equations:
(2x−3y= 7
3x+ 2y= 5
Solution
Step 1: Multiply the first equation by 2 and the second equation by 3 to elimi-
nate y.
(4x−6y= 14
9x+ 6y= 15
Step 2: Add the equations together to eliminate y.
13x= 29
Step 3: Solve for x.
x=29
13
13
Step 4: Substitute xback into one of the original equations to solve for y.
Let’s substitute into the first equation.
229
13−3y= 7
Step 5: Simplify and solve for y.
58
13 −3y= 7
58
13 −39
13 = 3y
19
13 = 3y
y=19
39
Step 6: The solution to the system of equations is x=29
13 and y=19
39 .
Question 16
Question
Solve the following system of equations:
(3x−2y= 7
2x+y=−5
Solution
Step 1: Solve the second equation for yin terms of x.
2x+y=−5
y=−2x−5
Step 2: Substitute y=−2x−5 into the first equation and solve for x.
3x−2(−2x−5) = 7
3x+ 4x+ 10 = 7
7x=−3
x=−3
7
14
Step 3: Substitute x=−3
7back into y=−2x−5 to find the value of y.
y=−2−3
7−5
y=6
7−5
y=−29
7
Therefore, the solution to the system of equations is x=−3
7and y=−29
7.
Question 17
Question
Solve the following system of equations:
(x2+y2= 29
x−y= 1
Solution
Step 1: Solve the second equation for xin terms of y.
x−y= 1
x=y+ 1
Step 2: Substitute xin terms of yinto the first equation.
(y+ 1)2+y2= 29
y2+ 2y+1+y2= 29
2y2+ 2y−28 = 0
y2+y−14 = 0
Step 3: Solve the quadratic equation y2+y−14 = 0.
(y+ 7)(y−2) = 0
y=−7,2
Step 4: Find the corresponding values of xusing x=y+ 1.
When y=−7, x =−6
When y= 2, x = 3
Thus, the solutions to the system of equations are (3,2) and (−6,−7).
15
Question 18
Question
Solve the following system of equations:
(3x+ 2y= 4
5x−3y=−7
Solution
Step 1: Multiply the first equation by 3 and the second equation by 2 to elimi-
nate y.
(9x+ 6y= 12
10x−6y=−14
Step 2: Add the two equations together to eliminate y.
19x=−2
Step 3: Solve for x.
x=−2
19
Step 4: Substitute xback into one of the original equations to solve for y.
Let’s use the first equation.
3−2
19 + 2y= 4
Step 5: Simplify and solve for y.
−6/19 + 2y= 4
2y= 4 + 6/19
y=38
19
So, the solution to the system of equations is x=−2
19 and y=38
19 .
Question 19
Question
Solve the following system of equations:
(3x+ 2y= 8
2x−4y=−5
16
Solution
Step 1: We will use the method of substitution to solve this system of equations.
Step 2: Let’s solve the first equation for x:
3x+ 2y= 8 =⇒x=8−2y
3
Step 3: Substitute xfrom the first equation into the second equation:
28−2y
3−4y=−5
Step 4: Simplify the equation:
16 −4y
3−4y=−5
16 −4y−12y=−15
16 −16y=−15
−16y=−31
y=31
16
Step 5: Substitute yback into the equation x=8−2y
3to find x:
x=8−231
16
3
x=8−62
16
3
x=8−62
16
3
x=8−31
8
3
x=64 −31
24
x=33
24
x=11
8
Step 6: Therefore, the solution to the system of equations is x=11
8and y=31
16 .
17
Question 20
Question
Solve the following system of equations:
(3x−2y= 7
6x−4y= 14
Solution
Step 1: We can start by multiplying the first equation by 2 to make the coeffi-
cients of ythe same in both equations:
(6x−4y= 14
6x−4y= 14
Step 2: Subtracting the second equation from the first, we get:
0=0
Step 3: Since we obtained a trivial equation, this means that the system of
equations have infinite solutions and the equations are dependent. This implies
that the two equations represent the same line in the plane.
Therefore, the system of equations has infinitely many solutions, and the
solution set can be written as follows:
{(x, y)|3x−2y= 7}
Question 21
Question
Solve the following system of equations:
(2x+ 3y= 7
4x−y= 6
Solution
Step 1: Solve the second equation for y:
4x−y= 6
y= 4x−6
18
Step 2: Substitute yin the first equation:
2x+ 3(4x−6) = 7
2x+ 12x−18 = 7
14x= 25
x=25
14
Step 3: Substitute xback into y= 4x−6 to find y:
y= 4 25
14−6
y=100
14 −84
14
y=16
14
y=8
7
Step 4: The solution to the system of equations is x=25
14 and y=8
7.
Question 22
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
14x= 22
x=22
14
x=11
7
19
Step 3: Substitute x=11
7back into y= 4x−5 to find y.
y= 4 11
7−5
y=44
7−5
y=44
7−35
7
y=9
7
Therefore, the solution to the system of equations is x=11
7and y=9
7.
Question 23
Question
Find the solution to the system of equations:
(3x+ 2y= 5
x2+xy +y2= 7
Solution
To solve this system of equations, we will first use the first equation to express
xin terms of y, then substitute it into the second equation.
Step 1: Solve the first equation for x.
3x+ 2y= 5
3x= 5 −2y
x=5−2y
3
Step 2: Substitute xinto the second equation.
5−2y
32
+5−2y
3y+y2= 7
(5 −2y)2
9+(5 −2y)y
3+y2= 7
25 −20y+ 4y2
9+5y−2y2
3+y2= 7
25 −20y+ 4y2+ 15y−6y2+ 9y2
9= 7
25 −5y+ 7y2
9= 7
25 −5y+ 7y2= 63
7y2−5y−38 = 0
20
Step 3: Solve the quadratic equation. Using the quadratic formula
y=−b±√b2−4ac
2a, we find:
y=5±p(−5)2−4(7)(−38)
2(7)
y=5±√25 + 1064
14
y=5±√1089
14
y=5±33
14
So, y=38
14 =19
7or y=−28
14 =−2.
Step 4: Find the corresponding xvalues for each yvalue. Using
x=5−2y
3: For y=19
7:
x=5−219
7
3
x=5−38
7
3
x=35 −38
21
x=−3
21
x=−1
7
For y=−2:
x=5−2(−2)
3
x=5+4
3
x=9
3
x= 3
Thus, the solutions to the system of equations are x=−1
7,y=19
7and
x= 3, y=−2.
21
Question 24
Question
Consider the following system of equations:
2x+ 3y+z= 4
x−y+ 2z= 3
3x+ 2y−z= 2
Determine whether the system has a unique solution, infinitely many solu-
tions, or no solution. If a solution exists, find the solution set.
Solution
Step 1: First, let’s write the system of equations in matrix form:
2 3 1
1−1 2
3 2 −1
x
y
z
=
4
3
2
Step 2: Next, we will find the determinant of the coefficient matrix on the
left side:
det
2 3 1
1−1 2
3 2 −1
=−22
Since the determinant is not zero, the system has a unique solution.
Step 3: To find the solution set, we will use Cramer’s Rule. Calculate the
determinants of the matrices formed by replacing the columns of the coefficient
matrix one at a time with the column vector on the right side:
Dx= det
4 3 1
3−1 2
2 2 −1
= 39
Dy= det
2 4 1
1 3 2
3 2 −1
=−28
Dz= det
234
1−1 3
322
= 18
Step 4: Finally, we can find the values of x,y, and zusing the formulas
x=Dx
det ,y=Dy
det , and z=Dz
det :
x=39
−22 =−39
22
22
y=−28
−22 =14
11
z=18
−22 =−9
11
Therefore, the unique solution to the system of equations is (x, y, z) =
−39
22 ,14
11 ,−9
11 .
Question 25
Question
Solve the following system of equations:
(2x−3y= 1
3x+ 4y= 4
Solution
Step 1: Begin by solving the first equation for xin terms of y.
2x−3y= 1 =⇒2x= 3y+ 1 =⇒x=3y+ 1
2
Step 2: Substitute xin the second equation with the expression 3y+1
2.
33y+ 1
2+ 4y= 4
Step 3: Simplify the expression and solve for y.
9y+ 3
2+4y= 4 =⇒9y+3+8y= 8 =⇒17y+3 = 8 =⇒17y= 5 =⇒y=5
17
Step 4: Substitute yback into the equation to solve for x.
x=3( 5
17 )+1
2=15
17 +17
17 =32
17
Step 5: Therefore, the solution to the system of equations is:
(x=32
17
y=5
17
Question 26
Question
Solve the following system of equations:
(2x−3y= 5
4x+ky = 9
23
Solution
Step 1: Let’s start by multiplying the first equation by 2 to eliminate xwhen
adding the equations.
(4x−6y= 10
4x+ky = 9
Step 2: Subtract the modified first equation from the second equation to
solve for y.
(4x+ky)−(4x−6y)=9−10
(4x−4x)+(ky + 6y) = −1
ky + 6y=−1
y(k+ 6) = −1
y=−1
k+ 6
Step 3: Substitute the expression for yback into the first equation and solve
for x.
2x−3−1
k+ 6= 5
2x+3
k+ 6 = 5
2x= 5 −3
k+ 6
2x=5(k+ 6) −3
k+ 6
2x=5k+ 30 −3
k+ 6
2x=5k+ 27
k+ 6
x=5k+ 27
2(k+ 6)
Therefore, the solution to the system of equations is:
(x=5k+27
2(k+6)
y=−1
k+6
Question 27
Question
Solve the following system of equations:
(2x+ 3y= 7
3x−2y= 8
24
Solution
Step 1: Let’s first multiply the first equation by 3 and the second equation by
2 to make the coefficients of ymatch for elimination.
(6x+ 9y= 21
6x−4y= 16
Step 2: Now, subtract the second equation from the first equation to elimi-
nate x:
13y= 5
Step 3: Solve for y:
y=5
13
Step 4: Substitute y=5
13 back into one of the original equations (let’s use
the first one) and solve for x:
2x+ 3 5
13= 7
2x+15
13 = 7
Step 5: Subtract 15
13 from both sides and solve for x:
2x=84
13
x=84
26
x=42
13
Step 6: The solution to the system of equations is x=42
13 and y=5
13 .
Question 28
Question
Solve the following system of equations:
(3x−2y= 5
2x+ 3y= 4
25
Solution
Step 1: Let’s use the elimination method to solve this system of equations. Our
goal is to eliminate one variable by adding or subtracting the equations.
Step 2: To eliminate y, multiply the first equation by 3 and the second
equation by 2:
(9x−6y= 15
4x+ 6y= 8
Step 3: Add the two equations:
13x= 23
Step 4: Solve for x:
x=23
13
Step 5: Substitute xback into the first equation to solve for y:
323
13−2y= 5
y=−4
13
Step 6: Therefore, the solution to the system of equations is:
(x=23
13
y=−4
13
Question 29
Question
Solve the following system of equations:
(3x+ 2y= 5
6x+ 4y= 10
Solution
Step 1: Multiply the first equation by 2 to create a system of equations with
the same coefficient for y.
(6x+ 4y= 10
6x+ 4y= 10
Step 2: Subtract one equation from the other to eliminate y.
(6x+ 4y)−(6x+ 4y) = 10 −10
0=0
26
Step 3: Since 0 = 0 is always true, the system has infinitely many solutions.
Thus, the solution set is all pairs of the form (x, 5−3x
2), where xis a real
number.
Question 30
Question
Solve the following system of equations:
(2x+ 3y= 7
3x+ 4y= 11
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+ 8y= 22
Step 2: Subtract the second equation from the first equation:
(6x+ 9y)−(6x+ 8y) = 21 −22
x+y=−1
Step 3: Substitute y=−1−xinto the first equation and solve for x:
2x+ 3(−1−x)=7
2x−3−3x= 7
−x−3=7
−x= 10
x=−10
Step 4: Substitute x=−10 into y=−1−xand solve for y:
y=−1−(−10)
y=−1 + 10
y= 9
Therefore, the solution to the system of equations is x=−10 and y= 9.
27
Question 31
Question
Solve the following system of equations:
3x+ 2y−z= 3
2x−y+z= 2
x+ 3y+ 2z= 4
Solution
Step 1: Let’s call the given system of equations (1), (2), and (3) respectively.
To solve the system, we will use the method of substitution.
Step 2: From equation (2), we can express zin terms of xand yas:
z= 2 −2x+y
Step 3: Substituting zin equations (1) and (3) with the expression 2−2x+y,
we get:
(3x+ 2y−(2 −2x+y) = 3
x+ 3y+ 2(2 −2x+y) = 4
Step 4: Simplifying equation (1) gives us:
5x+ 3y= 5
Step 5: Substituting equation (2) back into equation (3) gives us:
x+ 3y+ 4 −4x+ 2y= 4
Step 6: Simplifying equation (3) gives us:
−3x+ 5y= 0
Step 7: We now have a system of two equations:
(5x+ 3y= 5
−3x+ 5y= 0
Step 8: Solving the system of equations (4) and (5) simultaneously gives us
x= 5 and y=−5.
Step 9: Substituting x= 5 and y=−5 back into equation (2) to solve for z
gives us z= 2.
Step 10: Therefore, the solution to the system of equations is:
x= 5
y=−5
z= 2
28
Question 32
Question
Solve the following system of equations:
3x−2y+z= 4
2x+y−3z=−2
x−4y+ 5z= 8
Solution
Step 1: Multiply the second equation by 2 and the third equation by 3 to
eliminate xwhen adding/subtracting equations.
3x−2y+z= 4
4x+ 2y−6z=−4
3x−12y+ 15z= 24
Step 2: Subtract the first equation from the second equation to eliminate y.
3x−2y+z= 4
x−4z=−8
3x−12y+ 15z= 24
Step 3: Multiply the second equation by 3 and subtract it from the first
equation to eliminate x.
3x−2y+z= 4
x−4z=−8
3z= 36
Step 4: Solve the third equation to find z.
z= 12
Step 5: Substitute z= 12 back into the second equation to solve for x.
x−4(12) = −8
x−48 = −8
x= 40
Step 6: Substitute x= 40 and z= 12 back into the first equation to solve
for y.
3(40) −2y+ 12 = 4
29
120 −2y+ 12 = 4
−2y=−128
y= 64
Therefore, the solution to the system of equations is:
x= 40, y = 64, z = 12
Question 33
Question
Solve the following system of equations:
(2x+ 3y= 11
3x−2y= 1
Solution
Step 1: Multiply the first equation by 3 and the second equation by 2 to get:
(6x+ 9y= 33
6x−4y= 2
Step 2: Subtract the second equation from the first equation to eliminate x:
13y= 31
Step 3: Solve for y:
y=31
13 = 2.38
Step 4: Substitute yback into one of the original equations (let’s use the
first one):
2x+ 3(2.38) = 11
Step 5: Solve for x:
2x+ 7.14 = 11
Step 6: Solve for x:
2x= 3.86
Step 7: Solve for x:
x= 1.93
Therefore, the solution to the system of equations is x= 1.93 and y= 2.38.
30
Question 34
Question
Solve the following system of equations:
(2x+ 3y= 7
3x−4y=−4
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 the expression for xinto the second equation:
37−3y
2−4y=−4
Step 3: Solve the equation for y:
21 −9y
2−4y=−4
21 −9y−8y=−8
21 −17y=−8
−17y=−29
y=29
17
Step 4: Substitute the value of yback into the equation to solve for x:
x=7−329
17
2
x=7−87
17
2
x=7·17 −87
34
x=119 −87
34
x=32
34
x=16
17
Therefore, the solution to the system of equations is x=16
17 and y=29
17 .
31
Question 35
Question
Solve the following system of equations:
3x−y+ 2z= 5
x+ 4y−z=−3
2x−3y+ 2z= 6
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 yin terms of xand z.
3x−y+ 2z= 5
y= 3x+ 2z−5
Step 2: Next, substitute y= 3x+ 2z−5 into the second equation and solve
for x.
x+ 4(3x+ 2z−5) −z=−3
x+ 12x+ 8z−20 −z=−3
13x+ 7z= 17
13x=−7z+ 17
x=−7
13z+17
13
Step 3: Now substitute x=−7
13 z+17
13 back into y= 3x+ 2z−5 to find y
in terms of z.
y= 3 −7
13z+17
13+ 2z−5
y=−21
13z+51
13 + 2z−5
y=−21
13z+ 2z+51
13 −65
13
y=−7
13z−14
13
Step 4: Substitute x=−7
13 z+17
13 and y=−7
13 z−14
13 into the third equation
32
to solve for z.
2−7
13z+17
13−3−7
13z−14
13+ 2z= 6
−14
13z+34
13 +21
13z+42
13 + 2z= 6
7
13z+76
13 = 6
7
13z=38
13
z= 38
Step 5: Finally, substitute z= 38 back into x=−7
13 z+17
13 and y=
−7
13 z−14
13 to find the values of xand y.
x=−7
13(38) + 17
13 =−18
y=−7
13(38) −14
13 =−31
Therefore, the solution to the system of equations is x=−18, y=−31, and
z= 38.
33
