Solve by Substitution y=4x+1 , 2y-4x=14

$y = 4 x + 1$ , $2 y - 4 x = 14$

Step 1

Replace all occurrences of $y$ with $4 x + 1$ in each equation.

Replace all occurrences of $y$ in $2 y - 4 x = 14$ with $4 x + 1$ .

$2 (4 x + 1) - 4 x = 14$

$y = 4 x + 1$

Simplify the left side.

Simplify $2 (4 x + 1) - 4 x$ .

Simplify each term.

Apply the distributive property.

$2 (4 x) + 2 \cdot 1 - 4 x = 14$

$y = 4 x + 1$

Multiply $4$ by $2$ .

$8 x + 2 \cdot 1 - 4 x = 14$

$y = 4 x + 1$

Multiply $2$ by $1$ .

$8 x + 2 - 4 x = 14$

$y = 4 x + 1$

$8 x + 2 - 4 x = 14$

$y = 4 x + 1$

Subtract $4 x$ from $8 x$ .

$4 x + 2 = 14$

$y = 4 x + 1$

$4 x + 2 = 14$

$y = 4 x + 1$

$4 x + 2 = 14$

$y = 4 x + 1$

$4 x + 2 = 14$

$y = 4 x + 1$

Step 2

Solve for $x$ in $4 x + 2 = 14$ .

Move all terms not containing $x$ to the right side of the equation.

Subtract $2$ from both sides of the equation.

$4 x = 14 - 2$

$y = 4 x + 1$

Subtract $2$ from $14$ .

$4 x = 12$

$y = 4 x + 1$

$4 x = 12$

$y = 4 x + 1$

Divide each term in $4 x = 12$ by $4$ and simplify.

Divide each term in $4 x = 12$ by $4$ .

$\frac{4 x}{4} = \frac{12}{4}$

$y = 4 x + 1$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{12}{4}$

$y = 4 x + 1$

Divide $x$ by $1$ .

$x = \frac{12}{4}$

$y = 4 x + 1$

$x = \frac{12}{4}$

$y = 4 x + 1$

$x = \frac{12}{4}$

$y = 4 x + 1$

Simplify the right side.

Divide $12$ by $4$ .

$x = 3$

$y = 4 x + 1$

$x = 3$

$y = 4 x + 1$

$x = 3$

$y = 4 x + 1$

$x = 3$

$y = 4 x + 1$

Step 3

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

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

$y = 4 (3) + 1$

$x = 3$

Simplify the right side.

Simplify $4 (3) + 1$ .

Multiply $4$ by $3$ .

$y = 12 + 1$

$x = 3$

Add $12$ and $1$ .

$y = 13$

$x = 3$

$y = 13$

$x = 3$

$y = 13$

$x = 3$

$y = 13$

$x = 3$

Step 4

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(3, 13)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(3, 13)$

Equation Form:

$x = 3, y = 13$

Step 6

Solve by Substitution y=4x+1 , 2y-4x=14

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