QUESTION IMAGE
Question
practice
equations with variables on each side
solve each equation. check your solution. (examples 1, 2)
- ( a - 4 = 3a + 8 )
- ( 12b + 15 = 5b + 8 )
- ( 6c = 9 - c )
- ( 3d + 8 = 7 + 2d )
- ( 16 - 5z = -9z - 16 )
- ( 5w - 13 = 11 - 3w )
- ( \frac{1}{6}p + 8 = -\frac{1}{3}p + 14 )
- ( \frac{7}{16}y + 6 = 8 + \frac{3}{8}y )
- ( \frac{3}{4}x - 2 = \frac{5}{8}x + 3 )
- ( 5 - \frac{1}{2}r = \frac{3}{4}r - 10 )
- ( \frac{5}{9}s - 14 = \frac{7}{18}s - 16 )
- ( -7 + \frac{3}{14}t = \frac{1}{7}t - 8 )
- ( \frac{1}{4}v + 5 = \frac{2}{9}v + 7 )
- ( 3.4f + 8.5 = 2.8f + 6.1 )
Problem 1: $a - 4 = 3a + 8$
Step1: Isolate variable terms
$a - 3a = 8 + 4$
Step2: Simplify both sides
$-2a = 12$
Step3: Solve for $a$
$a = \frac{12}{-2} = -6$
Check:
Substitute $a=-6$: $-6 - 4 = 3(-6) + 8$ → $-10 = -18 + 8$ → $-10=-10$ (valid)
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Problem 3: $3c = 9 - c$
Step1: Isolate variable terms
$3c + c = 9$
Step2: Simplify and solve
$4c = 9$ → $c = \frac{9}{4} = 2.25$
Check:
Substitute $c=2.25$: $3(2.25) = 9 - 2.25$ → $6.75=6.75$ (valid)
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Problem 4: $3d + 8 = 14 - 2d$
Step1: Isolate variable terms
$3d + 2d = 14 - 8$
Step2: Simplify and solve
$5d = 6$ → $d = \frac{6}{5} = 1.2$
Check:
Substitute $d=1.2$: $3(1.2)+8=14-2(1.2)$ → $3.6+8=14-2.4$ → $11.6=11.6$ (valid)
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Problem 5: $36 - 5z = -9z - 16$
Step1: Isolate variable terms
$-5z + 9z = -16 - 36$
Step2: Simplify and solve
$4z = -52$ → $z = \frac{-52}{4} = -13$
Check:
Substitute $z=-13$: $36-5(-13)=-9(-13)-16$ → $36+65=117-16$ → $101=101$ (valid)
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Problem 6: $5w - 13 = 11 - 3w$
Step1: Isolate variable terms
$5w + 3w = 11 + 13$
Step2: Simplify and solve
$8w = 24$ → $w = \frac{24}{8} = 3$
Check:
Substitute $w=3$: $5(3)-13=11-3(3)$ → $15-13=11-9$ → $2=2$ (valid)
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Problem 7: $\frac{1}{6}p + 8 = -\frac{1}{3}p + 14$
Step1: Eliminate fractions (multiply by 6)
$p + 48 = -2p + 84$
Step2: Isolate variable terms
$p + 2p = 84 - 48$
Step3: Simplify and solve
$3p = 36$ → $p = 12$
Check:
Substitute $p=12$: $\frac{1}{6}(12)+8=-\frac{1}{3}(12)+14$ → $2+8=-4+14$ → $10=10$ (valid)
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Problem 8: $\frac{7}{10}y + 6 = 8 + \frac{3}{8}y$
Step1: Eliminate fractions (multiply by 40)
$28y + 240 = 320 + 15y$
Step2: Isolate variable terms
$28y - 15y = 320 - 240$
Step3: Simplify and solve
$13y = 80$ → $y = \frac{80}{13} \approx 6.15$
Check:
Substitute $y=\frac{80}{13}$: $\frac{7}{10}(\frac{80}{13})+6=8+\frac{3}{8}(\frac{80}{13})$ → $\frac{56}{13}+6=8+\frac{30}{13}$ → $\frac{56+78}{13}=\frac{104+30}{13}$ → $\frac{134}{13}=\frac{134}{13}$ (valid)
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Problem 9: $\frac{3}{4}x - 2 = \frac{5}{8}x + 3$
Step1: Eliminate fractions (multiply by 8)
$6x - 16 = 5x + 24$
Step2: Isolate variable terms
$6x - 5x = 24 + 16$
Step3: Solve for $x$
$x = 40$
Check:
Substitute $x=40$: $\frac{3}{4}(40)-2=\frac{5}{8}(40)+3$ → $30-2=25+3$ → $28=28$ (valid)
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Problem 10: $5 - \frac{1}{2}r = \frac{3}{4}r - 10$
Step1: Eliminate fractions (multiply by 4)
$20 - 2r = 3r - 40$
Step2: Isolate variable terms
$-2r - 3r = -40 - 20$
Step3: Simplify and solve
$-5r = -60$ → $r = 12$
Check:
Substitute $r=12$: $5-\frac{1}{2}(12)=\frac{3}{4}(12)-10$ → $5-6=9-10$ → $-1=-1$ (valid)
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Problem 11: $\frac{5}{9}s - 14 = \frac{7}{18}s - 16$
Step1: Eliminate fractions (multiply by 18)
$10s - 252 = 7s - 288$
Step2: Isolate variable terms
$10s - 7s = -288 + 252$
Step3: Simplify and solve
$3s = -36$ → $s = -12$
Check:
Substitute $s=-12$: $\frac{5}{9}(-12)-14=\frac{7}{18}(-12)-16$ → $-\frac{60}{9}-14=-\frac{84}{18}-16$ → $-\frac{20}{3}-14=-\frac{14}{3}-16$ → $\frac{-20-42}{3}=\frac{-14-48}{3}$ → $\frac{-62}{3}=\frac{-62}{3}$ (valid)
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Problem 12: $-7 + \frac{3}{14}t = \frac{1}{7}t - 8$
Step1: Eliminate fractions (multiply by 14)
$-98 + 3t = 2t - 112$
Step2: Isolate variable terms
$3t - 2t = -112 + 98$
Step3: Solve for $t$
$t = -14$
Check:
Substitute $t=-14$: $-7+\frac{3}{14}(-14)=\frac{1}{7}(-14)-8$ → $-7-3=-2-8$ → $-10=-10$ (valid)
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Problem 13: $\frac{1}{3}v + 5 = \frac{2}{9}v + 7$
Step1: Eliminate fractions (multiply by 9)
$3v + 45 =…
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- $a=-6$
- $c=\frac{9}{4}$
- $d=\frac{6}{5}$
- $z=-13$
- $w=3$
- $p=12$
- $y=\frac{80}{13}$
- $x=40$
- $r=12$
- $s=-12$
- $t=-14$
- $v=18$
- $f=-4$