Find the Points of Intersection 8x-2y=5 -12x+3y=7

$8 x - 2 y = 5$ $- 12 x + 3 y = 7$

Solve for $x$ in $8 x - 2 y = 5$ .

Add $2 y$ to both sides of the equation.

$8 x = 5 + 2 y$

$- 12 x + 3 y = 7$

Divide each term in $8 x = 5 + 2 y$ by $8$ and simplify.

Divide each term in $8 x = 5 + 2 y$ by $8$ .

$\frac{8 x}{8} = \frac{5}{8} + \frac{2 y}{8}$

$- 12 x + 3 y = 7$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 x}{8} = \frac{5}{8} + \frac{2 y}{8}$

$- 12 x + 3 y = 7$

Divide $x$ by $1$ .

$x = \frac{5}{8} + \frac{2 y}{8}$

$- 12 x + 3 y = 7$

$x = \frac{5}{8} + \frac{2 y}{8}$

$- 12 x + 3 y = 7$

$x = \frac{5}{8} + \frac{2 y}{8}$

$- 12 x + 3 y = 7$

Simplify the right side.

Cancel the common factor of $2$ and $8$ .

Factor $2$ out of $2 y$ .

$x = \frac{5}{8} + \frac{2 (y)}{8}$

$- 12 x + 3 y = 7$

Cancel the common factors.

Factor $2$ out of $8$ .

$x = \frac{5}{8} + \frac{2 y}{2 \cdot 4}$

$- 12 x + 3 y = 7$

Cancel the common factor.

$x = \frac{5}{8} + \frac{2 y}{2 \cdot 4}$

$- 12 x + 3 y = 7$

Rewrite the expression.

$x = \frac{5}{8} + \frac{y}{4}$

$- 12 x + 3 y = 7$

$x = \frac{5}{8} + \frac{y}{4}$

$- 12 x + 3 y = 7$

$x = \frac{5}{8} + \frac{y}{4}$

$- 12 x + 3 y = 7$

$x = \frac{5}{8} + \frac{y}{4}$

$- 12 x + 3 y = 7$

$x = \frac{5}{8} + \frac{y}{4}$

$- 12 x + 3 y = 7$

$x = \frac{5}{8} + \frac{y}{4}$

$- 12 x + 3 y = 7$

Replace all occurrences of $x$ with $\frac{5}{8} + \frac{y}{4}$ in each equation.

Replace all occurrences of $x$ in $- 12 x + 3 y = 7$ with $\frac{5}{8} + \frac{y}{4}$ .

$- 12 (\frac{5}{8} + \frac{y}{4}) + 3 y = 7$

$x = \frac{5}{8} + \frac{y}{4}$

Simplify the left side.

Simplify $- 12 (\frac{5}{8} + \frac{y}{4}) + 3 y$ .

Simplify each term.

Apply the distributive property.

$- 12 (\frac{5}{8}) - 12 \frac{y}{4} + 3 y = 7$

$x = \frac{5}{8} + \frac{y}{4}$

Cancel the common factor of $4$ .

Factor $4$ out of $- 12$ .

$4 (- 3) (\frac{5}{8}) - 12 \frac{y}{4} + 3 y = 7$

$x = \frac{5}{8} + \frac{y}{4}$

Factor $4$ out of $8$ .

$4 \cdot (- 3 \frac{5}{4 \cdot 2}) - 12 \frac{y}{4} + 3 y = 7$

$x = \frac{5}{8} + \frac{y}{4}$

Cancel the common factor.

$4 \cdot (- 3 \frac{5}{4 \cdot 2}) - 12 \frac{y}{4} + 3 y = 7$

$x = \frac{5}{8} + \frac{y}{4}$

Rewrite the expression.

$- 3 (\frac{5}{2}) - 12 \frac{y}{4} + 3 y = 7$

$x = \frac{5}{8} + \frac{y}{4}$

$- 3 (\frac{5}{2}) - 12 \frac{y}{4} + 3 y = 7$

$x = \frac{5}{8} + \frac{y}{4}$

Combine $- 3$ and $\frac{5}{2}$ .

$\frac{- 3 \cdot 5}{2} - 12 \frac{y}{4} + 3 y = 7$

$x = \frac{5}{8} + \frac{y}{4}$

Multiply $- 3$ by $5$ .

$\frac{- 15}{2} - 12 \frac{y}{4} + 3 y = 7$

$x = \frac{5}{8} + \frac{y}{4}$

Cancel the common factor of $4$ .

Factor $4$ out of $- 12$ .

$\frac{- 15}{2} + 4 (- 3) (\frac{y}{4}) + 3 y = 7$

$x = \frac{5}{8} + \frac{y}{4}$

Cancel the common factor.

$\frac{- 15}{2} + 4 \cdot (- 3 \frac{y}{4}) + 3 y = 7$

$x = \frac{5}{8} + \frac{y}{4}$

Rewrite the expression.

$\frac{- 15}{2} - 3 y + 3 y = 7$

$x = \frac{5}{8} + \frac{y}{4}$

$\frac{- 15}{2} - 3 y + 3 y = 7$

$x = \frac{5}{8} + \frac{y}{4}$

Move the negative in front of the fraction.

$- \frac{15}{2} - 3 y + 3 y = 7$

$x = \frac{5}{8} + \frac{y}{4}$

$- \frac{15}{2} - 3 y + 3 y = 7$

$x = \frac{5}{8} + \frac{y}{4}$

Combine the opposite terms in $- \frac{15}{2} - 3 y + 3 y$ .

Add $- 3 y$ and $3 y$ .

$- \frac{15}{2} + 0 = 7$

$x = \frac{5}{8} + \frac{y}{4}$

Add $- \frac{15}{2}$ and $0$ .

$- \frac{15}{2} = 7$

$x = \frac{5}{8} + \frac{y}{4}$

$- \frac{15}{2} = 7$

$x = \frac{5}{8} + \frac{y}{4}$

$- \frac{15}{2} = 7$

$x = \frac{5}{8} + \frac{y}{4}$

$- \frac{15}{2} = 7$

$x = \frac{5}{8} + \frac{y}{4}$

$- \frac{15}{2} = 7$

$x = \frac{5}{8} + \frac{y}{4}$

Since $- \frac{15}{2} = 7$ is not true, there is no solution.

No solution

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