A) Solving a system of equations by substitution with“y =” or “x =” equation.

Given a system of equations:

y = 2x – 5

x + 3y = 6

Combine the equations by replacing the ‘y’ with ‘2x – 5’ in the equation x + 3y = 6.

solve the new equation

x + 3(2x – 5) = 6Back substitute to find y Write the answer

x + 6x – 15 = 6y = 2x – 5 The solution

7x – 15 = 6y = 2(3) – 5 is (3,1)

7x – 15 + (15) = 6 + (15)y = 6 – 5

7x = 21y = 1

x = 3

B) Solving a system of equations by substitution without “y =” or “x =”.

Given a system of equations:

x + 2y = 4

3x + y = 7

Change one equation to y = or x =.

3x + y = 7

(–3x) + 3x + y = (–3x) + 7

y = –3x + 7

Combine the equations by replacing the ‘y’ with ‘–3x + 7’ in the equation x + 2y = 7.

solve the new equation

x + 2(–3x + 7) = 4Back substitute to find y Write the answer

x – 6x + 14 = 4 3x + y = 7 The solution

– 5x + 14 = 4 3(2) + y = 7 is (2,1)

–5x + 14 +(–14) = 4 + (–14) 6 + y = 7

– 5x = – 10 (–6) + 6 + y = (–6) + 7

x = 2 y = 1

Continue on back

C) Solving a system of equation using addition method (elimination method)

Given a system of equations

3x + y = 9

2x – y = 1If one variable can be eliminated by adding the equations, add the equations.

5x = 10Then, solve for the remaining variable.

x = 2

Back Substitutewrite the solution

3x + y = 9The solution is (2, 3)

3(2) + y = 9

6 + y = 9

(–6) + 6 + y = (–6) + 9

y = 3

D) Solving a system of equation using addition method (elimination method)

(multiply first to eliminate y)

Given a system of equations

4x + 2y = 10multiply by 312x + 6y = 30

x – 3y = 6multiply by 2 2x – 6y = 12 Add the equations and

14x = 42 solve for the remaining

x = 3 variable

Back Substitute

x – 3y = 6

3 – 3y = 6

(–3) + 3 – 3y = (–3) + 6Write the solution

–3y = 3The solution is (3, –1)

y = –1

E) Solving a system of equation using addition method (elimination method)

(multiply first to eliminate x)

Given a system of equations

4x + 2y = 10multiply by 1 4x + 2y = 10

x – 3y = 6multiply by –4–4x + 12y = –24 Add the equations and

14y = –14 solve for the remaining

y = –1 variable

Back Substitute

x – 3y = 6

x – 3(–1) = 6

x + 3 = 6Write the solution

x + 3 +(–3) = 6 + (–3)The solution is (3, –1)

x = 3
Solve each system of equations using the substitution method.

1. y = 6x - 17

3x - 6y = -30

2. y = 4x - 7

x + 8y = 76

3. x = y - 4

6x + 5y = -46

4. y = 2x

x - y = 4

5. 7x + y = -26

12x + 11y = 39

6. x - 7y = 62

3x + 2y = 2

Solve each system of equations using the addition method (elimination method)

7.x + y = 19

x - y = 7

8.-5x + 2y = -17

5x + 3y = -13

9.8x + 6y = 2

2x - y = 23

10.x + 5y = 19

x + 2y = 10

11.7x + 3y = 8

5x - 4y = -25

12.3x + y = 16

x + 7y = 32

13.5x + y = 16

6x + 5y = 61

14.5x + y = 37

x - 9y = -57

15. x - y = 25

5x - y = 69

16. 7x + 2y = -11

14x + 10y = 50

17.6x + 14y = 0

y = 2x – 17

Solve using a system of equations.

18.The sum of two numbers is 94. Their difference is 12. Find the numbers.

19.The sum of two numbers is 76. Their difference is 36. Find the numbers.

20.One number is 4 more than the other. Their sum is 8. Find the numbers.

21.One number is 8 more than the other. Their sum is 66. Find the numbers.

22.One number is 10 less than 4 times the other. Their sum is 195. Find the numbers.

23.If one number is 9 less than 7 times the other and their sum is 263, what are the numbers?

Key

Solve each system of equations using the substitution method.

1.ans: (4, 7)

2.ans: (4, 9)

3.ans: (-6, -2)

4.ans: (-4, -8)

5. ans: (-5, 9)

6.ans: (6, -8)

Solve each system of equations using the addition method (elimination method)

7.ans: (13, 6)

8.ans: (1,-6)

9.ans: (7, -9)

10.ans: (4, 3)

11.ans: (-1, 5)

12.ans: (4, 4)

13.ans: (1, 11)

14.ans: (6, 7)

15.ans: (11, -14)

16.ans: (-5, 12)

17.ans: (7, -3)

Solve each using a system of equations

18.The numbers are 53 and 41

19.The numbers are 56 and 20

20.The numbers are 2 and 6

21.The numbers are 29 and 37

22.The numbers are 41 and 154

23.The numbers are 34 and 229