# How do you find the point of intersection for x-4y=6 and 3x+4y=10?

Jul 30, 2015

The unique answer is the point $\left(x , y\right) = \left(4 , - \frac{1}{2}\right)$

#### Explanation:

Method 1 (Substitution): One way to solve this is to take the first equation of the system, $x - 4 y = 6$, and solve it for $x$ in terms of $y$ to get $x = 4 y + 6$. Then plug this into the second equation to make $3 x + 4 y = 10$ become $3 \left(4 y + 6\right) + 4 y = 10$, which simplifies to $12 y + 18 + 4 y = 10$, then $16 y = - 8$, then $y = - \frac{1}{2}$.

Now use the equation $x = 4 y + 6$ to find $x$: $x = 4 \cdot \left(- \frac{1}{2}\right) + 6 = - 2 + 6 = 4$.

The answer is the point $\left(x , y\right) = \left(4 , - \frac{1}{2}\right)$.

(BTW, all this work implies the only possible solution is $\left(x , y\right) = \left(4 , - \frac{1}{2}\right)$ (so if it is a solution, it is unique). You can confirm it truly is a solution by substitution back into the original system (checking your work proves it is a solution).)

Method 2 (Elimination): Notice that the $- 4 y$ in the first equation and the $4 y$ in the second equation will cancel if we add the two equations. Do so to get: $4 x = 16$ so $x = 4$. Then substitute this into either of the original equations to get, for instance $4 - 4 y = 6$ so $4 y = - 2$ and $y = - \frac{1}{2}$, resulting in $\left(x , y\right) = \left(4 , - \frac{1}{2}\right)$.

A picture of this situation is shown below. The equation $x - 4 y = 6$ is equivalent to $y = \frac{1}{4} x - \frac{3}{2}$ and is the red line with a slope of $\frac{1}{4}$ and a $y$-intercept of $- \frac{3}{2}$. The equation $3 x + 4 y = 10$ is equivalent to $y = - \frac{3}{4} x + \frac{5}{2}$ and is the blue line with a slope of $- \frac{3}{4}$ and a $y$-intercept of $\frac{5}{2}$. 