How do you solve x² + y² = 20 and x + y = 6 using substitution?

Sep 5, 2017

See a solution process below:

Explanation:

Step 1) Solve the second equation for $x$:

$x + y = 6$

$x + y - \textcolor{red}{y} = 6 - \textcolor{red}{y}$

$x + 0 = 6 - y$

$x = 6 - y$

Step 2) Substitute $\left(6 - y\right)$ for $x$ in the first equation and solve for $y$:

${x}^{2} + {y}^{2} = 20$ becomes:

${\left(6 - y\right)}^{2} + {y}^{2} = 20$

$36 - 12 y + {y}^{2} + {y}^{2} = 20$

$1 {y}^{2} + 1 {y}^{2} - 12 y + 36 = 20$

$\left(1 + 1\right) {y}^{2} - 12 y + 36 = 20$

$2 {y}^{2} - 12 y + 36 = 20$

$2 {y}^{2} - 12 y + 36 - \textcolor{red}{20} = 20 - \textcolor{red}{20}$

$2 {y}^{2} - 12 y + 16 = 0$

$\left(2 y - 8\right) \left(y - 2\right) = 0$

Solution 1 for $y$:

$2 y - 8 = 0$

$2 y - 8 + \textcolor{red}{8} = 0 + \textcolor{red}{8}$

$2 y - 0 = 8$

$2 y = 8$

$\frac{2 y}{\textcolor{red}{2}} = \frac{8}{\textcolor{red}{2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} y}{\cancel{\textcolor{red}{2}}} = 4$

$y = 4$

Solution 2 for $y$:

$y - 2 = 0$

$y - 2 + \textcolor{red}{2} = 0 + \textcolor{red}{2}$

$y - 0 = 2$

$y = 2$

Step 3) Substitute the values for $y$ into the solution to the second equation at the end of Step 1 and calculate $x$:

$x = 6 - y$ and $x = 6 - y$ becomes:

$x = 6 - 4$ and $x = 6 - 2$

$x = 2$ and $x = 4$

The Solutions Are:

$x = 2$ and $y = 4$ or $\left(2 , 4\right)$

AND

$x = 4$ and $y = 2$ or $\left(4 , 2\right)$