Solving Multi-Step Equations Quiz with six algebraic problems on two pages.

Here are the step-by-step solutions for each equation on the worksheet.

1) $3(a - \frac{2}{3}) = \frac{3}{4}a + 2\frac{1}{4}$

* Step 1: Distribute the 3 on the left side.
$3 \cdot a = 3a$
$3 \cdot (-\frac{2}{3}) = -2$
Equation becomes: $3a - 2 = \frac{3}{4}a + 2\frac{1}{4}$

* Step 2: Convert the mixed number to an improper fraction.
$2\frac{1}{4} = \frac{9}{4}$
Equation becomes: $3a - 2 = \frac{3}{4}a + \frac{9}{4}$

* Step 3: Move variables to one side and numbers to the other. Subtract $\frac{3}{4}a$ from both sides and add 2 to both sides.
$3a - \frac{3}{4}a = \frac{9}{4} + 2$

* Step 4: Simplify.
Left side: $3a$ is $\frac{12}{4}a$. So, $\frac{12}{4}a - \frac{3}{4}a = \frac{9}{4}a$.
Right side: $2$ is $\frac{8}{4}$. So, $\frac{9}{4} + \frac{8}{4} = \frac{17}{4}$.
Equation becomes: $\frac{9}{4}a = \frac{17}{4}$

* Step 5: Solve for $a$ by multiplying both sides by the reciprocal of $\frac{9}{4}$ (which is $\frac{4}{9}$).
$a = \frac{17}{4} \cdot \frac{4}{9}$
The 4s cancel out.
$a = \frac{17}{9}$ or $1\frac{8}{9}$

2) $\frac{z}{2} - \frac{3}{5} = -\frac{2}{3}z + \frac{1}{6}$

* Step 1: Find a common denominator for all fractions (2, 5, 3, 6) to clear them. The least common multiple is 30. Multiply every term by 30.
$30(\frac{z}{2}) - 30(\frac{3}{5}) = 30(-\frac{2}{3}z) + 30(\frac{1}{6})$

* Step 2: Simplify.
$15z - 18 = -20z + 5$

* Step 3: Move variables to the left and numbers to the right. Add $20z$ to both sides and add 18 to both sides.
$15z + 20z = 5 + 18$
$35z = 23$

* Step 4: Divide by 35.
$z = \frac{23}{35}$

3) $\frac{7}{4}x - 3 = 2 + \frac{9}{2}x$

* Step 1: Move variables to one side. Subtract $\frac{7}{4}x$ from both sides.
$-3 = 2 + \frac{9}{2}x - \frac{7}{4}x$

* Step 2: Find a common denominator for the x terms (4).
$\frac{9}{2}x = \frac{18}{4}x$
So, $\frac{18}{4}x - \frac{7}{4}x = \frac{11}{4}x$
Equation becomes: $-3 = 2 + \frac{11}{4}x$

* Step 3: Subtract 2 from both sides.
$-5 = \frac{11}{4}x$

* Step 4: Multiply by the reciprocal $\frac{4}{11}$.
$x = -5 \cdot \frac{4}{11}$
$x = -\frac{20}{11}$ or $-1\frac{9}{11}$

4) $\frac{3c + 8}{3} = \frac{1}{2} + \frac{c}{4}$

* Step 1: Clear denominators by multiplying everything by the least common multiple of 3, 2, and 4, which is 12.
$12 \cdot (\frac{3c + 8}{3}) = 12 \cdot (\frac{1}{2}) + 12 \cdot (\frac{c}{4})$

* Step 2: Simplify.
$4(3c + 8) = 6 + 3c$
$12c + 32 = 6 + 3c$

* Step 3: Subtract $3c$ from both sides and subtract 32 from both sides.
$12c - 3c = 6 - 32$
$9c = -26$

* Step 4: Divide by 9.
$c = -\frac{26}{9}$ or $-2\frac{8}{9}$

5) $\frac{1}{3} - \frac{2}{9}m = 15 + m$

* Step 1: Move $m$ terms to the right and numbers to the left. Add $\frac{2}{9}m$ to both sides and subtract 15 from both sides.
$\frac{1}{3} - 15 = m + \frac{2}{9}m$

* Step 2: Simplify both sides.
Left side: $15 = \frac{45}{3}$. So $\frac{1}{3} - \frac{45}{3} = -\frac{44}{3}$.
Right side: $m = \frac{9}{9}m$. So $\frac{9}{9}m + \frac{2}{9}m = \frac{11}{9}m$.
Equation becomes: $-\frac{44}{3} = \frac{11}{9}m$

* Step 3: Multiply by the reciprocal $\frac{9}{11}$.
$m = -\frac{44}{3} \cdot \frac{9}{11}$

* Step 4: Cross-cancel.
$44 \div 11 = 4$
$9 \div 3 = 3$
$m = -4 \cdot 3$
$m = -12$

6) $\frac{1}{2}(q + 1) = \frac{4}{3} - q$

* Step 1: Distribute the $\frac{1}{2}$.
$\frac{1}{2}q + \frac{1}{2} = \frac{4}{3} - q$

* Step 2: Add $q$ to both sides and subtract $\frac{1}{2}$ from both sides.
$\frac{1}{2}q + q = \frac{4}{3} - \frac{1}{2}$

* Step 3: Simplify.
Left side: $\frac{1}{2}q + \frac{2}{2}q = \frac{3}{2}q$
Right side: Common denominator is 6. $\frac{8}{6} - \frac{3}{6} = \frac{5}{6}$
Equation becomes: $\frac{3}{2}q = \frac{5}{6}$

* Step 4: Multiply by reciprocal $\frac{2}{3}$.
$q = \frac{5}{6} \cdot \frac{2}{3}$
$q = \frac{10}{18}$
Simplify fraction: $q = \frac{5}{9}$

7) $\frac{1}{6}r + 2 = 4\frac{1}{3}r + \frac{8}{3}$

* Step 1: Convert mixed number to improper fraction.
$4\frac{1}{3} = \frac{13}{3}$
Equation: $\frac{1}{6}r + 2 = \frac{13}{3}r + \frac{8}{3}$

* Step 2: Clear denominators by multiplying everything by 6 (LCM of 6 and 3).
$6(\frac{1}{6}r) + 6(2) = 6(\frac{13}{3}r) + 6(\frac{8}{3})$
$1r + 12 = 2(13r) + 2(8)$
$r + 12 = 26r + 16$

* Step 3: Subtract $r$ from both sides and subtract 16 from both sides.
$12 - 16 = 26r - r$
$-4 = 25r$

* Step 4: Divide by 25.
$r = -\frac{4}{25}$

8) $\frac{2}{3} - \frac{3}{2}y + \frac{1}{3}y + 4 = 0$

* Step 1: Combine like terms.
Numbers: $\frac{2}{3} + 4 = \frac{2}{3} + \frac{12}{3} = \frac{14}{3}$
Variables: $-\frac{3}{2}y + \frac{1}{3}y$. Common denominator is 6.
$-\frac{9}{6}y + \frac{2}{6}y = -\frac{7}{6}y$
Equation becomes: $\frac{14}{3} - \frac{7}{6}y = 0$

* Step 2: Add $\frac{7}{6}y$ to both sides.
$\frac{14}{3} = \frac{7}{6}y$

* Step 3: Multiply by reciprocal $\frac{6}{7}$.
$y = \frac{14}{3} \cdot \frac{6}{7}$

* Step 4: Cross-cancel.
$14 \div 7 = 2$
$6 \div 3 = 2$
$y = 2 \cdot 2$
$y = 4$

Final Answer:
1) $a = \frac{17}{9}$ (or $1\frac{8}{9}$)
2) $z = \frac{23}{35}$
3) $x = -\frac{20}{11}$ (or $-1\frac{9}{11}$)
4) $c = -\frac{26}{9}$ (or $-2\frac{8}{9}$)
5) $m = -12$
6) $q = \frac{5}{9}$
7) $r = -\frac{4}{25}$
8) $y = 4$