﻿ Solving Linear Systems by Elimination Worksheet | Problems & Solutions

# Solving Linear Systems by Elimination Worksheet

Solving Linear Systems by Elimination Worksheet
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1.
When the digits of a two-digit number are reversed, the new number is 7 more than twice the original number. Find the original number if the sum of its digits is 11.
 a. 29 b. 38 c. 92 d. 83

#### Solution:

Let t represent the tens digit and u represent the units digit.

Then, the original number is 10t + u and the new number is 10u + t.
[Two digit number = 10 × number in tens place + number in unit place.]

Tens digit + units digit = 11

t + u = 11 - - - (1)

New number = 7 + 2(original number).

10u + t = 7 + 2(10t + u)

10u + t = 7 + 20t + 2u

8u - 19t = 7 - - - (2)
[Simplify.]

8t + 8u = 88
19t - 8u = -7
--------------------
27t      =   81
[Multiply equation (1) by 8, equation (2) by - 1 and then add.]

t = 3

3 + u = 11
[Substitute the values.]

u = 8

Original number = 10t + u = 10(3) + 8 = 38

So, the original number is 38.

2.
A two-digit number is 9 less than the number obtained by reversing its digits. Find the original number if it is 5 times the sum of its digits
 a. 64 b. 46 c. 54 d. 45

#### Solution:

Let t represent the tens digit, u represent the units number.

Then, 10t + u represents the original number and 10u + t represents the number obtained by reversing the digits.
[Number = 10 × number in tens place + number in units place.]

5(tens digit + units digit) = original number

5(t + u) = 10t + u

- 5t + 4u = 0 - - - (1)
[Simplify.]

Reversed number - 9 = original number

10u + t - 9 = 10t + u

9u - 9t = 9

That is: u - t = 1 - - - (2)
[Divide throughout by 9.]

-5t + 4u = 0; - t + u = 1;

- 5t + 4u = 0
5t - 5 u = - 5
-------------------
- u = - 5
[Multiply second equation by - 5 and add to the first equation.]

u = 5
[Solve for u.]

- 5t + 4(5) = 0
[Substitute the values.]

t = 4

So, the original number is: 10t + u = 10(4) + 5 = 45.

3.
Find the solution to the system of equations 2$x$ + $y$ + $z$ = 4, $x$ - $y$ - $z$ = - 1, and 3$x$ + $y$ - $z$ = 3.
 a. $x$ = -1, $y$ = 1, $z$ = -1 b. $x$ = 1, $y$ = 1, $z$ = 1 c. $x$ = -1, $y$ = 1, $z$ = 1 d. $x$ = 1, $y$ = -1, $z$ = 1

#### Solution:

2x + y + z = 4  ---------(1)
x - y -  z = -1 ---------(2)
3x + y - z = 3  ---------(3)

Solve the equations (1) and (2):
2x +   y  +  z  = 4
-2x + 2y + 2z = 2
-------------------
3y + 3z = 6 ---------(4)
[Multiply equation (2) by -2 and add to the equation (1).]

Solve the equations (2) and (3):
-3x + 3y + 3z = 3
3x  +  y  -   z = 3
------------------
4y + 2z = 6 ---------(5)
[Multiply the second equation by -3 and add to the equation (3).]

Solve the equations (4) and (5):
-12y - 12z = -24
12y +  6z  = 18
-----------------
- 6z = - 6
[Multiply equation (4) by - 4, multiply equation (5) by 3 and then add.]

z = 1
[Solve for z.]

4y + 2(1) = 6 and hence y = 1
[Substitute z = 1 in equation (5) and solve for y.]

2x + (1) + (1) = 4 and hence x = 1
[Substitute y = 1, z = 1 in equation (1) and solve for x.]

The solution to the system is x = 1, y = 1, z = 1.

4.
Find the solution to the system of equations 4$x$ - $y$ + 5$z$ = 8, $x$ + $y$ + $z$ = 3, and 2$y$ + $z$ = 3.
 a. $x$ = -1, $y$ = -1, $z$ = 1 b. $x$ = 1, $y$ = 1, $z$ = -1 c. $x$ = 1, $y$ = 1, $z$ = 1 d. $x$ = -1, $y$ = 1, $z$ = -1

#### Solution:

4x - y + 5z = 8 -------(1)
x + yz  = 3 -------(2)
2y + z = 3 -------(3)

Solve the equations (1) and (2):
4x - y + 5z =  8
x + y +  z  =  3
------------------
5x       + 6z = 11 -------(4)

Solve the equations (2) and (3):
-2x - 2y - 2z = -6
2y +  z =  3
------------------
-2x          - z = -3

[Multiply equation (2) by -2 and add to the equation (3).]

2x + z = 3 -------(5)

Solve the equations (4) and (5):
5x + 6z =  11
-12x - 6z = -18
----------------
-7x        =   -7

[Multiply equation (5) by -6 and add to the equation (4).]

x = 1
[Solve of x.]

2(1) + z = 3 and hence z = 1
[Substitute x = 1 in equation (5) and solve for z.]

1 + y + 1 = 3 and hence y = 1
[Substitute x = 1, z = 1 in equation (2) and solve for y.]

The solution to the system is x = 1, y = 1, z = 1.

5.
Solve the system: 4$x$ - $y$ + $z$ = 5, $y$ + $z$ = - 1, and $x$ + $z$ = 4.
 a. $x$ = - 2, $y$ = - 7, $z$ = 6 b. $x$ = 2, $y$ = 5, $z$ = - 6 c. $x$ = -2, $y$ = - 7, $z$ = - 6 d. $x$ = 2, $y$ = -5, $z$ = 6

#### Solution:

4x - y + z =  5  ---------(1)
y + z = -1  ---------(2)
x       + z =  4  ---------(3)

Solve the equations (1) and (2):
4x - y + z =  5
y + z = -1
---------------------
4x    + 2z =  4 ---------(4)

Solve the equations (3) and (4):
- 4x - 4z = -16
4x + 2z =   4
----------------
- 2z = -12
[Multiply the equation (3) by - 4 and add to the equation (4).]

z = 6
[Solve for z.]

x + 6 = 4 and hence x = -2
[Substitute z = 6 in equation (3) and solve for x.]

y + 6 = -1 and hence y = - 7
[Substitute z = 6 in equation (2) and solve for y.]

The solution to the system is x = - 2, y = - 7 and z = 6.

6.
Solve: $x$ - $y$ + $z$ = 0, 3$x$ + $y$ + $z$ = 0, and 5$x$ + $z$ = 0.
 a. $x$ = 1, $y$ = 0, $z$ = 0 b. $x$ = 0, $y$ = 1, $z$ = 1 c. $x$ = 0, $y$ = 1, $z$ = 0 d. $x$ = 0, $y$ = 0, $z$ = 0

#### Solution:

x -  yz = 0 ---------(1)
3x + y + z = 0 ---------(2)
5x       + z = 0 ---------(3)

Solve the equations (1) and (2):
x -  y +   z = 0
3x + y +  z = 0
---------------------
4x      + 2z = 0 ---------(4)

Solve the equations (3) and (4):
-10x - 2z = 0
4x + 2z = 0
-----------------
-6x         = 0

[Multiple equation (3) by -2 and add to the equation (4).]

x = 0
[Substitute for x.]

4(0) + 2z = 0
[Substitute x = 0 in equation (4).]

z = 0
[Solve for z.]

0 - y + 0 = 0
[Substitute z = x = 0 in equation (1).]

y = 0
[Solve for y.]

The solution to the system is x = 0, y = 0 and z = 0.

7.
Solve: 7$x$ - $y$ + $z$ = 5, $x$ + 3$y$ = 4, and $z$ = - 1.
 a. $x$ = -1, $y$ = 1, $z$ = -1 b. $x$ = -1, $y$ = 1, $z$ = -1 c. $x$ = 1, $y$ = 1, $z$ = -1 d. $x$ = 1, $y$ = 1, $z$ = 1

#### Solution:

7x - y + z =  5  ----------(1)
x + 3y     =  4  ----------(2)
z = -1 ----------(3)

Solve the equations (1) and (3):
7x - y + z =  5
z = -1
--------------------
7x - y       =  6 ----------(4)

[Multiply equation (3) by -1 and add to the equation (3).]

Solve the equations (2) and (4):
x + 3y =   4
21x - 3y = 18
----------------
22x         = 22

[Multiply equation (4) by 3 and add to the equation (2).]

x = 1
[Solve for x.]

7(1) - y = 6
[Substitute x = 1 in equation (4).]

y = 1
[Solve for y.]

The solution to the system is x = 1, y = 1 and z = -1.

8.
Solve the system: 2$x$ - 3$y$ = -13, 4$x$ + $y$ = - 5
 a. $x$ = -2, $y$ = -3 b. $x$ = 2, $y$ = 3 c. $x$ = 2, $y$ = -3 d. $x$ = -2, $y$ = 3

#### Solution:

2x - 3y = - 13, 4x + y = - 5 are the two equations.

4x - 6y = -26
4x +  y = -5
---------------
-7y = - 21
[Multiply the first equation by 2 and then subtract the second equation.]

y = 3
[Solve for y.]

2x - 3(3) = -13
[Substitute y = 3 in the first equation.]

x = -2
[Solve for x.]

The solution to the given system is x = -2 and y = 3.

9.
Find the solution to the system of equations $x$ + 2$y$ = - 2; 3$x$ - 4$y$ = 9.
 a. $x$ = 1, $y$ = - $\frac{3}{2}$ b. $x$ = - 1, $y$ = - $\frac{3}{2}$ c. $x$ = -1, $y$ = $\frac{3}{2}$ d. $x$ = 1, $y$ = $\frac{3}{2}$

#### Solution:

x + 2y = -2, 3x - 4y = 9 are the two equations.

3x + 6y = -6
3x -  4y =  9
---------------
10y = -15
[Multiply the first equation by 3 and then subtract the second equation.]

y = - 3 / 2
[Solve for y.]

x + 2(- 3 / 2) = - 2
[Substitute y = - 32 in the first equation.]

x = 1
[Solve for x.]

The solution to the given system is x = 1, y = - 32.

10.
Solve the system: 2$x$ + 5$y$ = - 3, $x$ - 2$y$ = 1
 a. $x$ = $\frac{1}{9}$, $y$ = $\frac{5}{9}$ b. $x$ = - $\frac{1}{9}$, $y$ = $\frac{5}{9}$ c. $x$ = $\frac{1}{9}$, $y$ = - $\frac{5}{9}$ d. $x$ = - $\frac{1}{9}$, $y$ = - $\frac{5}{9}$

#### Solution:

2x + 5y = -3, x -  2y = 1 are the two equations.

2x + 5y = -3
2x -  4y =  2
---------------
9y = -5
[Multiply the second equation by 2 and then subtract from the first equation.]

y = - 59
[Solve for y.]

x - 2(- 59) = 1
[Substitute y = - 59 in the second equation.]

x = - 1 / 9

The solution to the given system is x = - 19, y = - 59.