Solve by Substitution 3x-5y=17 , 9x-15y=-4

$3 x - 5 y = 17$ , $9 x - 15 y = - 4$

Step 1

Solve for $x$ in $3 x - 5 y = 17$ .

Add $5 y$ to both sides of the equation.

$3 x = 17 + 5 y$

$9 x - 15 y = - 4$

Divide each term in $3 x = 17 + 5 y$ by $3$ and simplify.

Divide each term in $3 x = 17 + 5 y$ by $3$ .

$\frac{3 x}{3} = \frac{17}{3} + \frac{5 y}{3}$

$9 x - 15 y = - 4$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{17}{3} + \frac{5 y}{3}$

$9 x - 15 y = - 4$

Divide $x$ by $1$ .

$x = \frac{17}{3} + \frac{5 y}{3}$

$9 x - 15 y = - 4$

$x = \frac{17}{3} + \frac{5 y}{3}$

$9 x - 15 y = - 4$

$x = \frac{17}{3} + \frac{5 y}{3}$

$9 x - 15 y = - 4$

$x = \frac{17}{3} + \frac{5 y}{3}$

$9 x - 15 y = - 4$

$x = \frac{17}{3} + \frac{5 y}{3}$

$9 x - 15 y = - 4$

Step 2

Replace all occurrences of $x$ with $\frac{17}{3} + \frac{5 y}{3}$ in each equation.

Replace all occurrences of $x$ in $9 x - 15 y = - 4$ with $\frac{17}{3} + \frac{5 y}{3}$ .

$9 (\frac{17}{3} + \frac{5 y}{3}) - 15 y = - 4$

$x = \frac{17}{3} + \frac{5 y}{3}$

Simplify the left side.

Simplify $9 (\frac{17}{3} + \frac{5 y}{3}) - 15 y$ .

Simplify each term.

Apply the distributive property.

$9 (\frac{17}{3}) + 9 (\frac{5 y}{3}) - 15 y = - 4$

$x = \frac{17}{3} + \frac{5 y}{3}$

Cancel the common factor of $3$ .

Factor $3$ out of $9$ .

$3 (3) (\frac{17}{3}) + 9 (\frac{5 y}{3}) - 15 y = - 4$

$x = \frac{17}{3} + \frac{5 y}{3}$

Cancel the common factor.

$3 \cdot (3 (\frac{17}{3})) + 9 (\frac{5 y}{3}) - 15 y = - 4$

$x = \frac{17}{3} + \frac{5 y}{3}$

Rewrite the expression.

$3 \cdot 17 + 9 (\frac{5 y}{3}) - 15 y = - 4$

$x = \frac{17}{3} + \frac{5 y}{3}$

$3 \cdot 17 + 9 (\frac{5 y}{3}) - 15 y = - 4$

$x = \frac{17}{3} + \frac{5 y}{3}$

Multiply $3$ by $17$ .

$51 + 9 (\frac{5 y}{3}) - 15 y = - 4$

$x = \frac{17}{3} + \frac{5 y}{3}$

Cancel the common factor of $3$ .

Factor $3$ out of $9$ .

$51 + 3 (3) (\frac{5 y}{3}) - 15 y = - 4$

$x = \frac{17}{3} + \frac{5 y}{3}$

Cancel the common factor.

$51 + 3 \cdot (3 (\frac{5 y}{3})) - 15 y = - 4$

$x = \frac{17}{3} + \frac{5 y}{3}$

Rewrite the expression.

$51 + 3 (5 y) - 15 y = - 4$

$x = \frac{17}{3} + \frac{5 y}{3}$

$51 + 3 (5 y) - 15 y = - 4$

$x = \frac{17}{3} + \frac{5 y}{3}$

Multiply $5$ by $3$ .

$51 + 15 y - 15 y = - 4$

$x = \frac{17}{3} + \frac{5 y}{3}$

$51 + 15 y - 15 y = - 4$

$x = \frac{17}{3} + \frac{5 y}{3}$

Combine the opposite terms in $51 + 15 y - 15 y$ .

Subtract $15 y$ from $15 y$ .

$51 + 0 = - 4$

$x = \frac{17}{3} + \frac{5 y}{3}$

Add $51$ and $0$ .

$51 = - 4$

$x = \frac{17}{3} + \frac{5 y}{3}$

$51 = - 4$

$x = \frac{17}{3} + \frac{5 y}{3}$

$51 = - 4$

$x = \frac{17}{3} + \frac{5 y}{3}$

$51 = - 4$

$x = \frac{17}{3} + \frac{5 y}{3}$

$51 = - 4$

$x = \frac{17}{3} + \frac{5 y}{3}$

Step 3

Since $51 = - 4$ is not true, there is no solution.

No solution

Step 4

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