Solve by Substitution x+5y=4 and 3x+15y=-1

$x + 5 y = 4$ and $3 x + 15 y = - 1$

Step 1

Subtract $5 y$ from both sides of the equation.

$x = 4 - 5 y$

$3 x + 15 y = - 1$

Step 2

Replace all occurrences of $x$ with $4 - 5 y$ in each equation.

Replace all occurrences of $x$ in $3 x + 15 y = - 1$ with $4 - 5 y$ .

$3 (4 - 5 y) + 15 y = - 1$

$x = 4 - 5 y$

Simplify the left side.

Simplify $3 (4 - 5 y) + 15 y$ .

Simplify each term.

Apply the distributive property.

$3 \cdot 4 + 3 (- 5 y) + 15 y = - 1$

$x = 4 - 5 y$

Multiply $3$ by $4$ .

$12 + 3 (- 5 y) + 15 y = - 1$

$x = 4 - 5 y$

Multiply $- 5$ by $3$ .

$12 - 15 y + 15 y = - 1$

$x = 4 - 5 y$

$12 - 15 y + 15 y = - 1$

$x = 4 - 5 y$

Combine the opposite terms in $12 - 15 y + 15 y$ .

Add $- 15 y$ and $15 y$ .

$12 + 0 = - 1$

$x = 4 - 5 y$

Add $12$ and $0$ .

$12 = - 1$

$x = 4 - 5 y$

$12 = - 1$

$x = 4 - 5 y$

$12 = - 1$

$x = 4 - 5 y$

$12 = - 1$

$x = 4 - 5 y$

$12 = - 1$

$x = 4 - 5 y$

Step 3

Since $12 = - 1$ is not true, there is no solution.

No solution

Step 4