Solve by Substitution y=x-2 , 7y-4x=10

$y = x - 2$ , $7 y - 4 x = 10$

Step 1

Replace all occurrences of $y$ with $x - 2$ in each equation.

Replace all occurrences of $y$ in $7 y - 4 x = 10$ with $x - 2$ .

$7 (x - 2) - 4 x = 10$

$y = x - 2$

Simplify the left side.

Simplify $7 (x - 2) - 4 x$ .

Simplify each term.

Apply the distributive property.

$7 x + 7 \cdot - 2 - 4 x = 10$

$y = x - 2$

Multiply $7$ by $- 2$ .

$7 x - 14 - 4 x = 10$

$y = x - 2$

$7 x - 14 - 4 x = 10$

$y = x - 2$

Subtract $4 x$ from $7 x$ .

$3 x - 14 = 10$

$y = x - 2$

$3 x - 14 = 10$

$y = x - 2$

$3 x - 14 = 10$

$y = x - 2$

$3 x - 14 = 10$

$y = x - 2$

Step 2

Solve for $x$ in $3 x - 14 = 10$ .

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

Add $14$ to both sides of the equation.

$3 x = 10 + 14$

$y = x - 2$

Add $10$ and $14$ .

$3 x = 24$

$y = x - 2$

$3 x = 24$

$y = x - 2$

Divide each term in $3 x = 24$ by $3$ and simplify.

Divide each term in $3 x = 24$ by $3$ .

$\frac{3 x}{3} = \frac{24}{3}$

$y = x - 2$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{24}{3}$

$y = x - 2$

Divide $x$ by $1$ .

$x = \frac{24}{3}$

$y = x - 2$

$x = \frac{24}{3}$

$y = x - 2$

$x = \frac{24}{3}$

$y = x - 2$

Simplify the right side.

Divide $24$ by $3$ .

$x = 8$

$y = x - 2$

$x = 8$

$y = x - 2$

$x = 8$

$y = x - 2$

$x = 8$

$y = x - 2$

Step 3

Replace all occurrences of $x$ with $8$ in each equation.

Replace all occurrences of $x$ in $y = x - 2$ with $8$ .

$y = (8) - 2$

$x = 8$

Simplify the right side.

Subtract $2$ from $8$ .

$y = 6$

$x = 8$

$y = 6$

$x = 8$

$y = 6$

$x = 8$

Step 4

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

$(8, 6)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(8, 6)$

Equation Form:

$x = 8, y = 6$

Step 6

Solve by Substitution y=x-2 , 7y-4x=10

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