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# Solving Multi-Step Algebra Equations Practice Questions

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Updated on Sep 23, 2011

## Tips for Solving Multi-Step Equations

• There are at least two ways to show multiplication. You may be used to seeing multiplication shown with an × like this: 5 × 3 = 15. In equations, this becomes confusing. In algebra, the convention is to show multiplication with either a · like this: 5 · 3 = 15, or with parentheses like this: 5(3) = 15. Both conventions will be used in the answers, so you should get used to either one.
• Similarly, division can be shown using the standard division symbol ÷, as in 10 ÷ 2 = 5. Or it can be shown using a fraction bar like this: 10 ÷ 2 == 5. Use and get used to both.
• Check your answers before looking at the answer solutions. Just substitute the value you find for the variable and work each side of the equation as if it were a numerical expression. If the quantities you find are equal, your solution is correct.
• Dividing by a fraction is the same as multiplying by its reciprocal. For example, .
• To write an equation for a word problem, let the unknown quantity be equal to the variable. Then write the equation based on the information stated in the problem.

## Practice Questions

1. 4x + 7 = 10
2. 13x + 21 = 60
3. 3x – 8 = 16
4. 39 = 3a – 9
5. 4 = 4a + 20
6. 10a + 5 = 7
7. 0.3a + 0.25 = 1
8. 41 – 2m = 65
9. 4m – 14 = 50
10. 10s – 6 = 0
11. 8s – 7 = 41

Solve the following word problems by letting a variable equal the unknown quantity, making an equation from the information given, and then solving the equation.

1. A farmer is raising a hog that weighed 20 lbs. when he bought it. He expects it to gain 12 pounds per month. He will sell it when it weighs 200 lbs. How many months will it be before he will sell the animal?
2. Mary earns \$1.50 less than twice Bill's hourly wage. Mary earns \$12.50 per hour. What is Bill's hourly wage?
3. At year's end, a share of stock in Axon Corporation was worth \$37. This was \$8 less than three times its value at the beginning of the year. What was the price of a share of Axon stock at the beginning of the year?
4. Jennifer earned \$4,000 more than 1.5 times her former salary by changing jobs. She earned \$64,000 at her new job. What was her salary at her previous employment?
5. Twenty-five more girls than the number of boys participate in interscholastic sports at a local high school. If the number of girls participating is 105, how many boys participate?

### Answers

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.

Underlined equations show the simplified result.

 1. Subtract 7 from both sides of the equation. 4x + 7 – 7 = 11 – 7 Associate like terms. 4x + (7 – 7) = (11 – 7) Perform numerical operations. 4x + (0) = (4) Zero is the identity element for addition. 4x = 4 Divide both sides of the equation by 4. 4x ÷ 4 = 4 ÷ 4 x = 1 2.Subtract 21 from both sides of the equation. 13x + 21 – 21 = 60 – 21 Associate like terms. 13x + (21 – 21) = (60 – 21) Perform numerical operations. 13x + (0) = (39) Zero is the identity element for addition. 13x = 39 Divide both sides of the equation by 13. 13x ÷ 13 = 39 ÷ 13 x = 3 3.Add 8 to each side of the equation. 3x – 8 + 8 = 16 + 8 Change subtraction to addition and change the sign of the term that follows. 3x+ –8 + 8 = 16 + 8 Associate like terms. 3x + (–8 + 8) = 16 + 8 Perform numerical operations. 3x + (0) = 24 Zero is the identity element for addition. 3x = 24 Divide both sides of the equation by 3. 3x ÷ 3 = 24 ÷ 3 x = 8 4. Add 6 to each side of the equation. 5x – 6 + 6 = –26 + 6 Change subtraction to addition and change the sign of the term that follows. 5x + –6 + 6 = –26 + 6 Associate like terms. 5x + (–6 + 6) = –26 + 6 Perform numerical operations. 5x + (0) = –20 Zero is the identity element for addition. 5x = –20 Divide both sides of the equation by 5. 5x ÷ 5 = –20 ÷ 5 5. Subtract 4 from both sides of the equation. + 4 – 4 = 10 – 4 Associate like terms. + (4 – 4) = 10 – 4 Perform numerical operations. + (0) = 6 Zero is the identity element for addition. = 6 Multiply both sides of the equation by 3. 3() = 3(6) x = 18 6. Add 5 to each side of the equation. – 5 + 5 = 1 + 5 Change subtraction to addition and change the sign of the term that follows. + –5 + 5 = 1 + 5 Associate like terms. + (–5 + 5) = 1 + 5 Perform numerical operations. + (0) = 6 Zero is the identity element for addition. = 6 Multiply both sides of the equation by 7. 7() = 7(6) x = 42 7. Add 9 to each side of the equation. 39 + 9 = 3a – 9 + 9 Change subtraction to addition and change the sign of the term that follows. 39 + 9 = 3a + (–9 + 9) Associate like terms. 39 + 9 = 3a + (–9 + 9) Perform numerical operations. 48 = 3a + (0) Zero is the identity element for addition. 48 = 3a Divide both sides of the equation by 3. 48 ÷ 3 = 3a ÷ 3 16 = a 8. Subtract 20 from both sides of the equation. 4 – 20 = 4a + 20 – 20 Associate like terms. 4 – 20 = 4a + (20 – 20) Perform numerical operations –16 = 4a + (0) Zero is the identity element for addition. –16 = 4a Divide both sides of the equation by 4. 9. Subtract 5 from both sides of the equation. 10a + 5 – 5 = 7 – 5 Associate like terms. 10a + (5 – 5) = 7 – 5 Perform numerical operations. 10a + (0) = 2 Zero is the identity element for addition. 10a = 2 Divide both sides of the equation by 10. 10. Subtract 0.25 from both sides of the equation. 0.3a + 0.25 – 0.25 = 1 – 0.25 Associate like terms. 0.3a + (0.25 – 0.25) = 1 – 0.25 Perform numerical operations. 0.3a + (0) = 0.75 Zero is the identity element for addition. 0.3a = 0.75 Divide both sides of the equation by 0.3. Simplify the result. a = 2.5 11. Subtract 8 from both sides of the equation. m + 8 – 8 = 20 – 8 Associate like terms. m + (8 – 8) = 20 – 8 Perform numerical operations. m + (0) = 12 Zero is the identity element for addition. m = 12 Multiply both sides of the equation by the reciprocal of m = 18 12. Add 3 to both sides of the equation. 9 + 3 = m – 3 + 3 Change subtraction to addition and change the sign of the term that follows. 9 + 3 = m + –3 + 3 Associate like terms. 9 + 3 = m + (–3 + 3) Perform numerical operations. 12 = m + (0) Zero is the identity element for addition. 12 = Multiply both sides of the equation by the reciprocal of 16 = m 13. Subtract 41 from both sides of the equation. 41 – 41 – 2m = 65 – 41 Associate like terms. (41 – 41) – 2m = 65 – 41 Perform numerical operations. (0) – 2m = 24 Zero is the identity element for addition. –2m = 24 You can change the subtraction to addition and the sign of the term following to its opposite, which in this case is –2m. –2m = 24 Divide both sides of the equation by –2. Use the rules for operating with signed numbers. 14.Add 14 to each side of the equation. 4m – 14 + 14 = 50 + 14 Change subtraction to addition and change the sign of the term that follows. 4m + –14 + 14 = 50 + 14 Associate like terms. 4m + (–14 + 14) = 50 + 14 Perform numerical operations. 4m + (0) = 64 Zero is the identity element for addition. 4m = 64 Divide both sides of the equation by 4. m = 16 15. This equation presents a slightly different look. The variable in the numerator has a coefficient. There are two methods for solving. Subtract 16 from both sides of the equation. + 16 – 16 = 24 – 16 Associate like terms. + (16 – 16) = 24 – 16 Perform numerical operations. + (0) = 8 Zero is the identity element for addition. = 8 Multiply both sides of the equation by 5. Use rules for multiplying whole numbers and fractions. 2m = 40 Divide both sides by 2. m = 20 Or you can recognize that Then you would multiply by the reciprocal of the coefficient. m = 20 16. Add 6 to each side of the equation. 7m – 6 + 6 = –2.5 + 6 Change subtraction to addition and change the sign of the term that follows. 7m + –6 + 6 = –2.5 + 6 Associate like terms. 7m + (–6 + 6) = –2.5 + 6 Perform numerical operations. 7m + (0) = 3.5 Divide both sides of the equation by 7. m = 0.5 17. Add 6 to each side of the equation. 10s – 6 + 6 = 0 + 6 Change subtraction to addition and change the sign of the term that follows. 10s + –6 + 6 = 0 + 6 Associate like terms. 10s + (–6 + 6) = 0 + 6 Perform numerical operations. 10s + (0) = 6 Divide both sides of the equation by 10. Express the answer in the simplest form. 18. Subtract 2.7 from both sides of the equation. + 2.7 – 2.7 = 3 – 2.7 Associate like terms. + (2.7 – 2.7) = 3 – 2.7 Perform numerical operations. + (0) = 0.3 Multiply both sides of the equation by 4. 4() = 4(0.3) s = 1.2 19. Add 7 to each side of the equation. 8s – 7 + 7 = 41 + 7 Change subtraction to addition and change the sign of the term that follows. 8s + –7 + 7 = 41 + 7 Associate like terms. 8s + (–7 + 7) = 41 + 7 Perform numerical operations. 8s + (0) = 48 Divide both sides of the equation by 8. Express the answer in simplest form. s = 6 20. Subtract 25 from both sides of the equation. –55 – 25 = 25 – 25 – s Change subtraction to addition and change the sign of the term that follows. –55 + –25 = 25 + –25 + –s Associate like terms. (–55 + –25) = (25 + –25) + –s Perform numerical operations. –80 = (0) + –s Zero is the identity element for addition. –80 = –s You are to solve for s, but the term remaining is –s. If you multiply both sides by –1, you will be left with a +s or just s. –1(–80) = –1(–s) Use the rules for operating with signed numbers. 80 = s 21. Let x = the number of months. The number of months (x), times 12 (pounds per month), plus the starting weight (20), will be equal to 200 pounds. An equation that represents these words would be 12x + 20 = 200. Subtract 20 from both sides of the equation. 12x + 20 – 20 = 200 – 20 Associate like terms. 12x + (20 – 20) = 200 – 20 Perform numerical operations. 12x + (0) = 180 Divide both sides of the equation by 12. x = 15 The farmer would have to wait 15 months before selling his hog. 22. Let x = Bill's hourly wage. Then 2x less \$1.50 is equal to Mary's hourly wage. The equation representing the last statement would be 2x – 1.50 = 12.50. Add 1.50 to both sides of the equation. 2x – 1.50 + 1.50 = 12.50 + 1.50 Perform numerical operations. 2x = 14.00 Divide both sides of the equation by 2. x = 7.00 Bill's hourly wage is \$7.00 per hour. 23. Let x = the share price at the beginning of the year. The statements tell us that if we multiply the share price at the beginning of the year by 3 and then subtract \$8, it will equal \$37. An equation that represents this amount is 3x – 8 = 37. Add 8 to both sides of the equation. 3x – 8 + 8 = 37 + 8 Perform numerical operations. 3x = 45 Divide both sides of the equation by 3. x = 15 One share of Axon costs \$15 at the beginning of the year. 24. Let x = her previous salary. The statements tell us that \$64,000 is equal to 1.5 times x plus \$4,000. An algebraic equation to represent this statement is 64,000 = 1.5x + 4,000. Subtract 4,000 from both sides of the equation. 64,000 – 4,000 = 1.5x + 4,000 – 4,000 Perform numerical operations. 60,000 = 1.5x Divide both sides of the equation by 1.5. 40,000 = x Jennifer's former salary was \$40,000 per year. 25. Let x = the number of boys who participate in interscholastic sports. The question tells us that the number of boys plus 25 is equal to the number of girls who participate. An equation that represents this statement is x + 25 = 105. Subtract 25 from both sides of the equation. x + 25 – 25 = 105 – 25 Perform numerical operations. x = 80 Multiply by the reciprocal of x = 120 The number of boys who participate is 120.
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