two-variables linear equations (ax+by=c) Math Worksheets, Math ... - Free Printable

Let's solve each system of two-variable linear equations step by step. We'll use either the elimination or substitution method depending on which is more efficient for each pair.

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Problem 1:

$$
\begin{align*}
(1)\quad & 8x + 7y = 99 \\
(2)\quad & 8x + 2y = 74 \\
\end{align*}
$$

Step 1: Subtract equation (2) from equation (1):
$$
(8x + 7y) - (8x + 2y) = 99 - 74 \\
5y = 25 \Rightarrow y = 5
$$

Step 2: Plug $ y = 5 $ into equation (2):
$$
8x + 2(5) = 74 \Rightarrow 8x + 10 = 74 \Rightarrow 8x = 64 \Rightarrow x = 8
$$

✔ Solution: $ x = 8, y = 5 $

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Problem 2:

$$
\begin{align*}
(1)\quad & 4x + 3y = 30 \\
(2)\quad & 6x + 8y = 52 \\
\end{align*}
$$

We'll eliminate one variable. Let’s make the coefficients of $ x $ the same.

Multiply (1) by 3 and (2) by 2:
$$
(1)\times3: \quad 12x + 9y = 90 \\
(2)\times2: \quad 12x + 16y = 104 \\
$$

Now subtract:
$$
(12x + 16y) - (12x + 9y) = 104 - 90 \Rightarrow 7y = 14 \Rightarrow y = 2
$$

Plug $ y = 2 $ into (1):
$$
4x + 3(2) = 30 \Rightarrow 4x + 6 = 30 \Rightarrow 4x = 24 \Rightarrow x = 6
$$

✔ Solution: $ x = 6, y = 2 $

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Problem 3:

$$
\begin{align*}
(1)\quad & 6x - 3y = 36 \\
(2)\quad & 4x - 6y = 16 \\
\end{align*}
$$

Simplify both equations:

Divide (1) by 3: $ 2x - y = 12 $ → (1a)

Divide (2) by 2: $ 2x - 3y = 8 $ → (2a)

Now subtract (2a) from (1a):
$$
(2x - y) - (2x - 3y) = 12 - 8 \Rightarrow 2y = 4 \Rightarrow y = 2
$$

Plug into (1a): $ 2x - 2 = 12 \Rightarrow 2x = 14 \Rightarrow x = 7 $

✔ Solution: $ x = 7, y = 2 $

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Problem 4:

$$
\begin{align*}
(1)\quad & 2x - 7y = -33 \\
(2)\quad & 8x - 8y = 8 \\
\end{align*}
$$

Simplify (2): divide by 8 → $ x - y = 1 $ → (2a) → $ x = y + 1 $

Substitute into (1):
$$
2(y + 1) - 7y = -33 \Rightarrow 2y + 2 - 7y = -33 \Rightarrow -5y = -35 \Rightarrow y = 7
$$

Then $ x = y + 1 = 8 $

✔ Solution: $ x = 8, y = 7 $

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Problem 5:

$$
\begin{align*}
(1)\quad & 5x - 6y = -2 \\
(2)\quad & 6x - 2y = 8 \\
\end{align*}
$$

Use elimination. Multiply (1) by 1 and (2) by 3 to eliminate $ y $:

(1): $ 5x - 6y = -2 $
(2)×3: $ 18x - 6y = 24 $

Now subtract:
$$
(18x - 6y) - (5x - 6y) = 24 - (-2) \Rightarrow 13x = 26 \Rightarrow x = 2
$$

Plug into (2): $ 6(2) - 2y = 8 \Rightarrow 12 - 2y = 8 \Rightarrow -2y = -4 \Rightarrow y = 2 $

✔ Solution: $ x = 2, y = 2 $

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Problem 6:

$$
\begin{align*}
(1)\quad & 8x - 8y = -16 \\
(2)\quad & 4x + 8y = 76 \\
\end{align*}
$$

Add equations directly:
$$
(8x - 8y) + (4x + 8y) = -16 + 76 \Rightarrow 12x = 60 \Rightarrow x = 5
$$

Plug into (1): $ 8(5) - 8y = -16 \Rightarrow 40 - 8y = -16 \Rightarrow -8y = -56 \Rightarrow y = 7 $

✔ Solution: $ x = 5, y = 7 $

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Problem 7:

$$
\begin{align*}
(1)\quad & 3x + 5y = 43 \\
(2)\quad & 4x - 6y = -44 \\
\end{align*}
$$

Use elimination. Make coefficients of $ x $ match.

LCM of 3 and 4 is 12.

Multiply (1) by 4: $ 12x + 20y = 172 $
Multiply (2) by 3: $ 12x - 18y = -132 $

Subtract:
$$
(12x + 20y) - (12x - 18y) = 172 - (-132) \Rightarrow 38y = 304 \Rightarrow y = 8
$$

Plug into (1): $ 3x + 5(8) = 43 \Rightarrow 3x + 40 = 43 \Rightarrow 3x = 3 \Rightarrow x = 1 $

✔ Solution: $ x = 1, y = 8 $

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Problem 8:

$$
\begin{align*}
(1)\quad & 4x + 6y = 62 \\
(2)\quad & 7x - 3y = 14 \\
\end{align*}
$$

Make coefficients of $ y $ match. Multiply (2) by 2:
$$
(2)\times2: \quad 14x - 6y = 28
$$

Now add to (1):
$$
(4x + 6y) + (14x - 6y) = 62 + 28 \Rightarrow 18x = 90 \Rightarrow x = 5
$$

Plug into (1): $ 4(5) + 6y = 62 \Rightarrow 20 + 6y = 62 \Rightarrow 6y = 42 \Rightarrow y = 7 $

✔ Solution: $ x = 5, y = 7 $

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✔ Final Answers:

| Problem | Solution |
|--------|----------|
| 1 | $ x = 8, y = 5 $ |
| 2 | $ x = 6, y = 2 $ |
| 3 | $ x = 7, y = 2 $ |
| 4 | $ x = 8, y = 7 $ |
| 5 | $ x = 2, y = 2 $ |
| 6 | $ x = 5, y = 7 $ |
| 7 | $ x = 1, y = 8 $ |
| 8 | $ x = 5, y = 7 $ |

Let me know if you'd like these solutions checked graphically or via substitution!