Solve by Substitution y=3x+1 , 4y=12x+4

$y = 3 x + 1$ , $4 y = 12 x + 4$

Step 1

Replace all occurrences of $y$ with $3 x + 1$ in each equation.

Replace all occurrences of $y$ in $4 y = 12 x + 4$ with $3 x + 1$ .

$4 (3 x + 1) = 12 x + 4$

$y = 3 x + 1$

Simplify the left side.

Simplify $4 (3 x + 1)$ .

Apply the distributive property.

$4 (3 x) + 4 \cdot 1 = 12 x + 4$

$y = 3 x + 1$

Multiply.

Multiply $3$ by $4$ .

$12 x + 4 \cdot 1 = 12 x + 4$

$y = 3 x + 1$

Multiply $4$ by $1$ .

$12 x + 4 = 12 x + 4$

$y = 3 x + 1$

$12 x + 4 = 12 x + 4$

$y = 3 x + 1$

$12 x + 4 = 12 x + 4$

$y = 3 x + 1$

$12 x + 4 = 12 x + 4$

$y = 3 x + 1$

$12 x + 4 = 12 x + 4$

$y = 3 x + 1$

Step 2

Remove any equations from the system that are always true.

$y = 3 x + 1$

Step 3

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