# How do you solve the following system using substitution?: 3x - 2 y = 5, x+ 4 y = 4

Mar 18, 2018

The point of intersection is $\left(2 , \frac{1}{2}\right)$.

#### Explanation:

Solve system of equations:

$\text{Equation 1} :$ $3 x - 2 y = 5$

$\text{Equation 2} :$ $x + 4 y = 4$

The two equations are linear equations in standard form. The solved values for $x$ and $y$ will be the point of intersection of the two lines.

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Solve Equation 2 for $x$.

Subtract $4 y$ from both sides of the equation.

$x = 4 - 4 y$

Substitute $\left(4 - 4 y\right)$ for $x$ in Equation 1 and solve for $y$.

$3 \left(4 - 4 y\right) - 2 y = 5$

Expand.

$12 - 12 y - 2 y = 5$

Simplify.

$12 - 14 y = 5$

Subtract $12$ from both sides of the equation.

$- 14 y = 5 - 12$

Simplify.

$- 14 y = - 7$

Divide both sides by $- 14$.

$y = \frac{- 7}{- 14}$

Simplify.

$y = \frac{1}{2}$

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Substitute $\frac{1}{2}$ for $y$ in Equation 2 and solve for $x$.

$x + 4 \left(\frac{1}{2}\right) = 4$

$x + {\cancel{4}}^{2} \left(\frac{1}{\cancel{2}} ^ 1\right) = 4$

Simplify.

$x + 2 = 4$

Subtract $2$ from both sides.

$x = 4 - 2$

Simplify.

$x = 2$

The point of intersection is $\left(2 , \frac{1}{2}\right)$.