# How do you solve the simultaneous equations x^2 + y^2=29 and y-x=3?

Jul 19, 2015

Use the second equation to provide an expression for $y$ in terms of $x$ to substitute into the first equation to give a quadratic equation in $x$.

#### Explanation:

First add $x$ to both sides of the second equation to get:

$y = x + 3$

Then substitute this expression for $y$ into the first equation to get:

$29 = {x}^{2} + {\left(x + 3\right)}^{2} = 2 {x}^{2} + 6 x + 9$

Subtract $29$ from both ends to get:

$0 = 2 {x}^{2} + 6 x - 20$

Divide both sides by $2$ to get:

$0 = {x}^{2} + 3 x - 10 = \left(x + 5\right) \left(x - 2\right)$

So $x = 2$ or $x = - 5$

If $x = 2$ then $y = x + 3 = 5$.

If $x = - 5$ then $y = x + 3 = - 2$

So the two solutions $\left(x , y\right)$ are $\left(2 , 5\right)$ and $\left(- 5 , - 2\right)$

Jul 19, 2015

$\left(x = - 5 \mathmr{and} y = - 2\right) \mathmr{and} \left(x = 2 \mathmr{and} y = 5\right)$

#### Explanation:

Since you have both ${x}^{2} + {y}^{2} = 29$ and $y - x = 3$,

You want to combine these two equations into one equation with a single variable, solve it and then solve for the other variable. An example on how to do this goes like this:

$y - x = 3 \rightarrow y = x + 3$ and we have ${y}^{2} = {x}^{2} + 6 x + 9$

Since ${x}^{2} + {y}^{2} = 29$, substitute the expression for ${y}^{2}$ into this:

$2 {x}^{2} + 6 x + 9 = 29$, so $2 {x}^{2} + 6 x - 20 = 0$.

We can solve for $x$ using the quadratic formula:
$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 2 \cdot \left(- 20\right)}}{2 \cdot 2} = - \frac{3}{4} \pm \frac{1}{4} \sqrt{196} = \frac{- 6 \pm 14}{4}$
So $x = - 5$ or $x = 2$.

Since $y = x + 3$, this gives $\left(x = - 5 \mathmr{and} y = - 2\right) \mathmr{and} \left(x = 2 \mathmr{and} y = 5\right)$.