Solve by Addition/Elimination 2x+4y=10 3x-4y=5

$2 x + 4 y = 10$ $3 x - 4 y = 5$

Add the two equations together to eliminate $y$ from the system.

Divide each term in $5 x = 15$ by $5$ and simplify.

Divide each term in $5 x = 15$ by $5$ .

$\frac{5 x}{5} = \frac{15}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{15}{5}$

Divide $x$ by $1$ .

$x = \frac{15}{5}$

Simplify the right side.

Divide $15$ by $5$ .

$x = 3$

Substitute the value found for $x$ into one of the original equations, then solve for $y$ .

Substitute the value found for $x$ into one of the original equations to solve for $y$ .

$2 (3) + 4 y = 10$

Multiply $2$ by $3$ .

$6 + 4 y = 10$

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

Subtract $6$ from both sides of the equation.

$4 y = 10 - 6$

Subtract $6$ from $10$ .

$4 y = 4$

Divide each term in $4 y = 4$ by $4$ and simplify.

Divide each term in $4 y = 4$ by $4$ .

$\frac{4 y}{4} = \frac{4}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 y}{4} = \frac{4}{4}$

Divide $y$ by $1$ .

$y = \frac{4}{4}$

Simplify the right side.

Divide $4$ by $4$ .

$y = 1$

The solution to the independent system of equations can be represented as a point.

$(3, 1)$

The result can be shown in multiple forms.

Point Form:

$(3, 1)$

Equation Form:

$x = 3, y = 1$