# How do you solve the simultaneous equations y=3x - 5 and x^2 + y^2 = 25?

Mar 28, 2018

Solution pairs of $\left(x , y\right)$ are $\left(0 , - 5\right)$ and $\left(3 , 4\right)$

#### Explanation:

After plugging first equation to second one,

${x}^{2} + {\left(3 x - 5\right)}^{2} = 25$

${x}^{2} + 9 {x}^{2} - 30 x + 25 = 25$

$10 {x}^{2} - 30 x + 25 = 25$

$10 {x}^{2} - 30 x = 0$

${x}^{2} - 3 x = 0$

$x \cdot \left(x - 3\right) = 0$

Hence ${x}_{1} = 0$ and ${x}_{2} = 3$. So ${y}_{1} = - 5$ and ${y}_{2} = 4$. Thus, solution pairs of $\left(x , y\right)$ are $\left(0 , - 5\right)$ and $\left(3 , 4\right)$