Linear Equations Worksheet featuring four algebraic equations to solve, ideal for student practice.

Let’s solve each equation one by one, step by step.

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Problem 1a:

Equation:
\[
\frac{3(p + 8)}{1} = \frac{-6(p + 6)}{-3}
\]

Step 1: Simplify both sides.

Left side:
\[
\frac{3(p + 8)}{1} = 3(p + 8)
\]

Right side:
\[
\frac{-6(p + 6)}{-3} = \frac{-6}{-3}(p + 6) = 2(p + 6)
\]

Now the equation is:
\[
3(p + 8) = 2(p + 6)
\]

Step 2: Distribute (multiply out the parentheses).

Left: \(3p + 24\)
Right: \(2p + 12\)

So:
\[
3p + 24 = 2p + 12
\]

Step 3: Get all variable terms on one side and constants on the other.

Subtract \(2p\) from both sides:
\[
p + 24 = 12
\]

Subtract 24 from both sides:
\[
p = 12 - 24 = -12
\]

✔ Check: Plug p = -12 back in.

Left: \(3(-12 + 8) = 3(-4) = -12\)
Right: \(2(-12 + 6) = 2(-6) = -12\) → Matches!

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Problem 1b:

Equation:
\[
\frac{-6s}{-1} = \frac{-2s + 7}{-7}
\]

Step 1: Simplify left side.

\[
\frac{-6s}{-1} = 6s
\]

Right side stays as is for now:
\[
\frac{-2s + 7}{-7}
\]

We can split the fraction:
\[
= \frac{-2s}{-7} + \frac{7}{-7} = \frac{2s}{7} - 1
\]

So equation becomes:
\[
6s = \frac{2s}{7} - 1
\]

Step 2: Eliminate fractions. Multiply every term by 7.

\[
7 \cdot 6s = 7 \cdot \left(\frac{2s}{7}\right) - 7 \cdot 1
\]

→ \(42s = 2s - 7\)

Step 3: Move variables to one side.

Subtract \(2s\) from both sides:
\[
40s = -7
\]

Divide by 40:
\[
s = -\frac{7}{40}
\]

✔ Check: Plug s = -7/40 into original.

Left: \(\frac{-6*(-7/40)}{-1} = \frac{42/40}{-1} = -42/40 = -21/20\)

Right: \(\frac{-2*(-7/40) + 7}{-7} = \frac{14/40 + 7}{-7} = \frac{14/40 + 280/40}{-7} = \frac{294/40}{-7} = \frac{294}{40} \cdot \frac{1}{-7} = \frac{294}{-280} = -\frac{294 ÷ 14}{280 ÷ 14} = -\frac{21}{20}\) → Matches!

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Problem 2a:

Equation:
\[
\frac{3v}{3} - 3v = -5
\]

Step 1: Simplify \(\frac{3v}{3} = v\)

So:
\[
v - 3v = -5
\]

→ \(-2v = -5\)

Step 2: Divide both sides by -2:

\[
v = \frac{-5}{-2} = \frac{5}{2}
\]

✔ Check: Plug v = 5/2

Left: \(\frac{3*(5/2)}{3} - 3*(5/2) = \frac{15/2}{3} - 15/2 = (15/2)*(1/3) - 15/2 = 5/2 - 15/2 = -10/2 = -5\) → Matches!

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Problem 2b:

Equation:
\[
-3(-4 - b) = 5 - \frac{8b}{4}
\]

Step 1: Simplify both sides.

Left: distribute -3
→ \(-3 * (-4) + (-3)*(-b) = 12 + 3b\)

Right: simplify fraction → \(\frac{8b}{4} = 2b\), so right side is \(5 - 2b\)

Equation now:
\[
12 + 3b = 5 - 2b
\]

Step 2: Move variables to one side, constants to other.

Add \(2b\) to both sides:
\[
12 + 5b = 5
\]

Subtract 12 from both sides:
\[
5b = 5 - 12 = -7
\]

Divide by 5:
\[
b = -\frac{7}{5}
\]

✔ Check: Plug b = -7/5

Left: \(-3(-4 - (-7/5)) = -3(-4 + 7/5) = -3(-20/5 + 7/5) = -3(-13/5) = 39/5\)

Right: \(5 - \frac{8*(-7/5)}{4} = 5 - \frac{-56/5}{4} = 5 - (-56/5 * 1/4) = 5 + 56/20 = 5 + 14/5 = 25/5 + 14/5 = 39/5\) → Matches!

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Final Answer:

1a. \( p = -12 \)
1b. \( s = -\frac{7}{40} \)
2a. \( v = \frac{5}{2} \)
2b. \( b = -\frac{7}{5} \)