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EXAMPLE 1

ANSWER. The solution is 2 . Check by substituting 2 for x in the original equation. EXAMPLE 1. Solve an equation with variables on both sides. Solve 7 – 8 x = 4 x – 17. 7 – 8 x = 4 x – 17. Write original equation. 7 – 8 x + 8 x = 4 x – 17 + 8 x. Add 8 x to each side.

michael-crosby
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EXAMPLE 1

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  1. ANSWER The solution is 2. Check by substituting 2 for xin the original equation. EXAMPLE 1 Solve an equation with variables on both sides Solve7 – 8x= 4x– 17. 7 – 8x = 4x – 17 Write original equation. 7 – 8x + 8x = 4x – 17 + 8x Add 8xto each side. 7 = 12x – 17 Simplify each side. 24 = 12x Add 17 to each side. 2 = x Divide each side by 12.

  2. ? 7 – 8(2) = 4(2) – 17 ? 29 =4(2) –17 EXAMPLE 1 Solve an equation with variables on both sides CHECK 7 – 8x= 4x– 17 Write original equation. Substitute 2 for x. Simplify left side. –9=–9 Simplify right side. Solution checks.

  3. (16x + 60). Solve 9x–5= (16x + 60). 9x – 5= 1 1 4 4 EXAMPLE 2 Solve an equation with grouping symbols Write original equation. 9x – 5 = 4x + 15 Distributive property 5x – 5 = 15 Subtract 4xfrom each side. 5x =20 Add 5 to each side. x = 4 Divide each side by 5.

  4. ANSWER The solution is 3. Check by substituting 3 for min the original equation. for Examples 1 and 2 GUIDED PRACTICE 1. 24 – 3m= 5m. 24 – 3m= 5m Write original equation. 24 – 3m+ 3m= 5m+ 3m Add 3mto each side. 24 = 8m Simplify each side. 3 = m Divide each side by 8.

  5. ? 24 – 3(3) = 5(3) ? 15=5(3) for Examples 1 and 2 GUIDED PRACTICE CHECK 24 – 3m= 5m Write original equation. Substitute 3 for m. Simplify left side. 15=15 Simplify right side. Solution checks.

  6. ANSWER The solution is 9. Check by substituting 9 for cin the original equation. for Examples 1 and 2 GUIDED PRACTICE 2. 20 +c= 4c – 7. 20 +c= 4c – 7 Write original equation. 20 +c – c= 4c – c – 7 Subtract cfrom each side. 20 = 3c – 7 Simplify each side. 27 = 3c Add 7 to each side. 9 = c Divide each side by 3.

  7. ? 20 + 9 = 4(9) – 7 ? 29=4(9) – 7 for Examples 1 and 2 GUIDED PRACTICE CHECK 20 +c= 4c – 7 Write original equation. Substitute 9 for c. Simplify left side. 29=29 Simplify right side. Solution checks.

  8. ANSWER The solution is –8. Check by substituting –8 for kin the original equation. for Examples 1 and 2 GUIDED PRACTICE 3. 9 – 3k= 17k – 2k. 9 – 3k= 17k – 2k Write original equation. 9 – 3k + 3k= 17k – 2k + 3k Add 3kto each side. 9 = 17k + k Simplify each side. – 8 = k Subtract 17 from each side.

  9. ? 9 –3(– 8) = 17 – (– 8)2 ? 33=17 – (– 8)2 for Examples 1 and 2 GUIDED PRACTICE CHECK 9 – 3k= 17– 2k Write original equation. Substitute –8 for k. Simplify left side. 33=33 Simplify right side. Solution checks.

  10. ANSWER The solution is 6. Check by substituting 6 for zin the original equation. for Examples 1 and 2 GUIDED PRACTICE 4. 5z– 2= 2(3z – 4). 5z– 2= 2(3z – 4) Write original equation. 5z– 2= 6z – 8 Distributive property. – z – 2 = – 8 Subtract 6z from each side. z = 6 Add zto each side.

  11. ? 5(6) – 2=2(3(6)– 4) ? 28=2(3(6)– 4) for Examples 1 and 2 GUIDED PRACTICE CHECK 5z– 2 =2(3z – 4) Write original equation. Substitute 6 for z. Simplify left side. 28=28 Simplify right side. Solution checks.

  12. ANSWER The solution is 2. Check by substituting 2 for ain the original equation. for Examples 1 and 2 GUIDED PRACTICE 5. 3 – 4a= 5(a – 3). 3 – 4a= 5(a – 3) Write original equation. 3 – 4a = 5a – 15 Distributive property. 3– 9a = – 15 Subtract 5a from each side. – 9a = – 18 Subtract 3 from each side. a = 2 Divide each side by –9.

  13. ? 3 – 4(2)=5(2– 3) ? – 5=5(2 –3) for Examples 1 and 2 GUIDED PRACTICE CHECK 3 – 4a=5(a – 3) Write original equation. Substitute 2 for a. Simplify left side. – 5=– 5 Simplify right side. Solution checks.

  14. 6. (6y + 15). (6y + 15). 8y–6= 8y–6= 2 2 3 3 ANSWER The solution is 4. Check by substituting 4 for yin the original equation. for Examples 1 and 2 GUIDED PRACTICE Write original equation. 8y – 6 = 4y +10 Distributive property 4y – 6 = 10 Subtract 4yfrom each side. 4y =16 Add 6 to each side. y = 4 Divide each side by 4.

  15. (6y + 15). 8y–6= ? 8(4) –6= (6(4)+ 15) ? 26=(6(4)+ 15) 2 2 2 3 3 3 for Examples 1 and 2 GUIDED PRACTICE CHECK Write original equation. Substitute 4 for y. Simplify left side. 26=26 Simplify right side. Solution checks.

  16. EXAMPLE 3 Solve a real-world problem CAR SALES A car dealership sold 78 new cars and 67 used cars this year. The number of new cars sold by the dealership has been increasing by 6 cars each year. The number of used cars sold by the dealership has been decreasing by 4 cars each year. If these trends continue, in how many years will the number of new cars sold be twice the number of used cars sold?

  17. 78 6x 67 (– 4 x) ) + + = 2 ( EXAMPLE 3 Solve a real-world problem SOLUTION Let xrepresent the number of years from now. So, 6x represents the increase in the number of new cars sold over xyears and – 4xrepresents the decrease in the number of used cars sold over x years. Write a verbal model.

  18. ANSWER The number of new cars sold will be twice the number of used cars sold in 4 years. EXAMPLE 3 Solve a real-world problem 78 + 6x =2(67 – 4x) Write equation. 78 + 6x = 134 – 8x Distributive property 78 + 14x =134 Add 8xto each side. 14x = 56 Subtract 78 from each side. x = 4 Divide each side by 14.

  19. EXAMPLE 3 Solve a real-world problem CHECK You can use a table to check your answer.

  20. 7. WHAT IF? Suppose the car dealership sold 50 new cars this year instead of 78. In how many years will the number of new cars sold be twice the number of used cars sold? for Example 3 GUIDED PRACTICE

  21. 50 6x 67 (–4x) ) + + = 2 ( for Example 3 GUIDED PRACTICE SOLUTION Let xrepresent the number of years from now. So, 6x represents the increase in the number of new cars sold over xyears and – 4xrepresents the decrease in the number of used cars sold over x years. Write a verbal model.

  22. ANSWER The number of new cars sold will be twice the number of used cars sold in 6 years. for Example 3 GUIDED PRACTICE 50 + 6x =2(67 +(– 4x)) Write equation. 50 + 6x = 134 – 8x Distributive property 50 + 14x =134 Add 8xto each side. 14x = 84 Subtract 50 from each side. x = 6 Divide each side by 14.

  23. EXAMPLE 4 Identify the number of solutions of an equation Solve the equation, if possible. a. 3x = 3(x + 4) b.2x + 10 = 2(x + 5) SOLUTION a. 3x = 3(x + 4) Original equation 3x = 3x + 12 Distributive property The equation 3x=3x+12 is not true because the number 3xcannot be equal to 12 more than itself. So, the equation has no solution. This can be demonstrated by continuing to solve the equation.

  24. 0 = 12 ANSWER The statement 0 = 12 is not true, so the equation has no solution. EXAMPLE 4 Identify the number of solutions of an equation 3x – 3x= 3x + 12 – 3x Subtract 3xfrom each side. Simplify.

  25. ANSWER Notice that the statement 2x + 10 = 2x + 10 is true for all values of x.So, the equation is an identity, and the solution is all real numbers. EXAMPLE 1 EXAMPLE 4 Identify the number of solutions of an equation b.2x + 10 = 2(x + 5) Original equation 2x + 10 = 2x + 10 Distributive property

  26. for Example 4 GUIDED PRACTICE 8. 9z + 12= 9(z + 3) SOLUTION 9z + 12= 9(z + 3) Original equation 9z + 12= 9z + 27 Distributive property The equation 9z + 12=9z + 27 is not true because the number 9z + 12cannot be equal to 27 more than itself. So, the equation has no solution. This can be demonstrated by continuing to solve the equation.

  27. 12 = 27 ANSWER The statement 12 = 27 is not true, so the equation has no solution. for Example 4 GUIDED PRACTICE 9z – 9z+ 12= 9z – 9z + 27 Subtract 9zfrom each side. Simplify.

  28. ANSWER w= 0 for Example 4 GUIDED PRACTICE 9. 7w + 1= 8w + 1 SOLUTION – w + 1= 1 Subtract 8w from each side. – w = 0 Subtract 1 from each side.

  29. ANSWER The statement 6a +6 = 6a +6 is true for all values of a. So, the equation is an identity, and the solution is all real numbers. for Example 4 GUIDED PRACTICE 10. 3(2a + 2)= 2(3a + 3) SOLUTION 3(2a + 2)= 2(3a + 3) Original equation 6a +6= 6a +6 Distributive property

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