PROBLEM SET
In the following exercises, solve the linear equation using the general strategy.
1. [latex]4\left(y+7\right)=64[/latex]
2. [latex]9=3\left(x - 3\right)[/latex]
3. [latex]14\left(y - 6\right)=-42[/latex]
4. [latex]-7\left(3n+4\right)=14[/latex]
5. [latex]8\left(3+3\text{p}\right)=0[/latex]
6. [latex]\Large\frac{3}{5}\normalsize\left(10x - 5\right)=27[/latex]
7. [latex]4\left(2.5v - 0.6\right)=7.6[/latex]
8. [latex]0.5\left(16m+34\right)=-15[/latex]
9. [latex]-\left(t - 8\right)=17[/latex]
10. [latex]8\left(6b - 7\right)+23=63[/latex]
11. [latex]13+2\left(m - 4\right)=17[/latex]
12. [latex]-9+6\left(5-k\right)=12[/latex]
13. [latex]18-\left(9r+7\right)=-16[/latex]
14. [latex]18 - 2\left(y - 3\right)=32[/latex]
15. [latex]3\left(4n - 1\right)-2=8n+3[/latex]
16. [latex]5\left(x - 4\right)-4x=14[/latex]
17. [latex]5+6\left(3s - 5\right)=-3+2\left(8s - 1\right)[/latex]
18. [latex]4\left(x - 1\right)-8=6\left(3x - 2\right)-7[/latex]
everyday math
Making a fence
Jovani has a fence around the rectangular garden in his backyard. The perimeter of the fence is [latex]150[/latex] feet. The length is [latex]15[/latex] feet more than the width. Find the width, [latex]w[/latex], by solving the equation [latex]150=2\left(w+15\right)+2w[/latex].
Coins Rhonda has [latex]{$1.90}[/latex] in nickels and dimes. The number of dimes is one less than twice the number of nickels. Find the number of nickels, [latex]n[/latex], by solving the equation [latex]0.05n+0.10\left(2n - 1\right)=1.90[/latex].
In the following exercises, solve the equation by clearing the fractions.
19. [latex]\Large\frac{1}{4}\normalsize x-\Large\frac{1}{2}\normalsize =-\Large\frac{3}{4}[/latex]
20. [latex]\Large\frac{5}{6}\normalsize y-\Large\frac{2}{3}\normalsize =-\Large\frac{3}{2}[/latex]
21. [latex]\Large\frac{1}{2}\normalsize a+\Large\frac{3}{8}\normalsize =\Large\frac{3}{4}[/latex]
22. [latex]2=\Large\frac{1}{3}\normalsize x-\Large\frac{1}{2}\normalsize x+\Large\frac{2}{3}\normalsize x[/latex]
23. [latex]\Large\frac{1}{4}\normalsize m-\Large\frac{4}{5}\normalsize m+\Large\frac{1}{2}\normalsize m=-1[/latex]
24. [latex]x+\Large\frac{1}{2}\normalsize =\Large\frac{2}{3}\normalsize x-\Large\frac{1}{2}[/latex]
25. [latex]\Large\frac{1}{3}\normalsize w+\Large\frac{5}{4}\normalsize =w-\Large\frac{1}{4}[/latex]
26. [latex]\Large\frac{1}{2}\normalsize x-\Large\frac{1}{4}\normalsize =\Large\frac{1}{12}\normalsize x+\Large\frac{1}{6}[/latex]
27. [latex]\Large\frac{1}{3}\normalsize b+\Large\frac{1}{5}\normalsize =\Large\frac{2}{5}\normalsize b-\Large\frac{3}{5}[/latex]
28. [latex]1=\Large\frac{1}{6}\normalsize\left(12x - 6\right)[/latex]
29. [latex]\Large\frac{1}{2}\normalsize\left(x+4\right)=\Large\frac{3}{4}[/latex]
In the following exercises, solve the equation by clearing the decimals.
30. [latex]0.9x - 1.25=0.75x+1.75[/latex]
31. [latex]0.10d+0.25\left(d+5\right)=4.05[/latex]
32. [latex]0.05\left(q - 5\right)+0.25q=3.05[/latex]
Everyday math
Coins Taylor has [latex]{$2.00}[/latex] in dimes and pennies. The number of pennies is [latex]2[/latex] more than the number of dimes. Solve the equation [latex]0.10d+0.01\left(d+2\right)=2[/latex] for [latex]d[/latex], the number of dimes.