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# Solving Systems of Equations Algebraically Practice Questions

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Updated on Sep 23, 2011

## Introduction

There is a faster way to solve systems of equations than graphing and finding the solution point. You can solve systems of equations using algebraic methods. The two methods you will practice here are called the elimination method and the substitution method. You will be using the skills you have practiced in the chapters on working with algebraic expressions, combining like terms, and solving equations.

In the elimination method, you will transform one or both of the two equations in the system so that when you add the two equations together, one of the variables will be eliminated. Then you solve the remaining equation for the remaining variable. When you find a numerical value for the remaining variable, you just substitute the found value into one of the equations and solve for the other variable.

In the substitution method, you will transform one of the equations so that one variable is expressed in terms of the other. Then you will eliminate the variable by substituting into the other equation and solve. When you find a numerical value for one variable, use it in one of the two equations to determine the value of the remaining variable.

One method is not better than the other. But you may find that you will begin to see which equations, because of their structure, lend themselves to one method over the other. Practice will help you decide.

## Tips for Solving Systems of Equations Algebraically

When using the elimination method, first make a plan to determine which variable you will eliminate from the system. Then transform the equation or equations so that you will get the result you want.

Express your solution as a coordinate point or in the form (x,y), or as variables such as x = 2 and y = 4.

## Practice Questions

Use the elimination method to solve the following systems of equations.

1. x + y = 4
2. 2xy = –1

3. 3x + 4y = 17
4. x + 2y = 1

5. 7x + 3y = 11
6. 2x + y = 3

7. 0.5x + 5y = 28
8. 3xy = 13

9. 3(x + y) = 18
10. 5x + y = –2

11. x + 2y = 11
12. 2xy = 17

13. 5x + 8y = 25
14. 3x – 15 = y

15. 6y + 3x = 30
16. 2y + 6x = 0

17. 3x = 5 – 7y
18. 2y = x – 6

19. 3x + y = 20
20. + 10 = y

21. 2x + 7y = 45
22. 3x + 4y = 22

23. 3x – 5y = –21
24. 2(2yx) = 16

25. x + y = 12
26. 2xy = 21

Use the substitution method to solve the following systems of equations.

1. y = 5x
2. 4x + 5y = 87

3. x + y = 3
4. 3x + 101 = 7y

5. 5x + y = 3.6
6. y + 21x = 8.4

7. 8xy = 0
8. 10x + y = 27

9. = y + 2
10. 2x – 4y = 32

11. y + 3x = 0
12. y – 3x = 24

13. 5x + y = 20
14. 3x = y + 1

15. 2x + y = 2 – 5y
16. xy = 5

17. 3x + 2y = 12

18. x + 6y = 11
19. x – 3 = 2y

20. 4y + 31 = 3x
21. y + 10 = 3x

22. 2(2 – x) = 3y – 2
23. 3x + 9 = 4(5 – y)

Numerical expressions in parentheses like this [ ] are operations performed on only part of the original expression The operations performed within these symbols are intended to show how to evaluate the various terms that make up the entire expression

Expressions with parentheses that look like this ( ) contain either numerical substitutions or expressions that are part of a numerical expression Once a single number appears within these parentheses, the parentheses are no longer needed and need not be used the next time the entire expression is written

When two pair of parentheses appear side by side like this ( )( ), it means that the expressions within are to be multiplied

Sometimes parentheses appear within other parentheses in numerical or algebraic expressions Regardless of what symbol is used, ( ), { }, or [ ], perform operations in the innermost parentheses first and work outward

The underlined ordered pair is the solution Be aware that you may have used a different method of elimination to arrive at the correct answer

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