# How do you solve x^2+y^2=109 and x-y= - 7 using substitution?

Mar 28, 2018

$\therefore \text{The Solution Set=} \left\{\left(x , y\right) = \left(3 , 10\right) , \left(- 10 , - 3\right)\right\}$.

#### Explanation:

$x - y = - 7 \Rightarrow x = y - 7$.

Sub.ing in, ${x}^{2} + {y}^{2} = 109$, we get,

${\left(y - 7\right)}^{2} + {y}^{2} = 109 , \mathmr{and} ,$

$2 {y}^{2} - 14 y + 49 - 109 = 0 , i . e . ,$

${y}^{2} - 7 y - 30 = 0$.

Using $10 \times 3 = 30 , \mathmr{and} , 10 - 3 = 7$, we get,

$\underline{{y}^{2} - 10 y} + \underline{3 y - 30} = 0$.

$\therefore y \left(y - 10\right) + 3 \left(y - 10\right) = 0$.

$\therefore \left(y - 10\right) \left(y + 3\right) = 0$.

$\therefore y = 10 , \mathmr{and} , y = - 3$.

 y=10, &, x=y-7 rArr x=3,and, y=-3 rArr x=-10.

These roots satisfy the given eqns.

$\therefore \text{The Solution Set=} \left\{\left(x , y\right) = \left(3 , 10\right) , \left(- 10 , - 3\right)\right\}$.

Mar 28, 2018

$x = - 10 , y = - 3 \mathmr{and} x = 3 , y = 10$

#### Explanation:

Rearrange $x - y = - 7$ so $x = y - 7$
Now substitute "$y - 7$ " for the $x$ in the first equation.

${x}^{2} + {y}^{2} = 109$ becomes ${\left(y - 7\right)}^{2} + {y}^{2} = 109$
Expand the bracket:
${y}^{2} - 7 y - 7 y + 49 + {y}^{2} = 109$

Collect like terms:
$2 {y}^{2} - 14 y + 49 = 109$

subtract 109 from both sides:
$2 {y}^{2} - 14 y - 60 = 0$

Factorise:
$\left(2 y + 6\right) \left(y - 10\right) = 0$

so $2 y + 6 = 0$ or $y - 10 = 0$

then $y = - 3$ or $y = 10$

Now put these two values into $x - y = - 7$
when $y = - 3$,
$x - \left(- 3\right) = - 7$
$x + 3 = - 7$
$x = - 10$

when $y = 10$,
$x - 10 = - 7$
$x = 3$

$x = - 10 \mathmr{and} y = - 3$
and
$x = 3 \mathmr{and} y = 10$

Mar 28, 2018

Express the implicit relation in terms of one variable only. See method below.

#### Explanation:

You have two equations:

${x}^{2} + {y}^{2} = 109$

$x - y = - 7$

You can express $x$ in terms of $y$ or express $y$ in terms of $x$. Either way is fine.

Since $x - y = - 7$, that means $y = x + 7$

Using this substitution we can write ${x}^{2} + {\left(x + 7\right)}^{2} = 109$

This will give us a quadratic expression where we can solve for $x$.

${x}^{2} + {\left(x + 7\right)}^{2} = {x}^{2} + \left({x}^{2} + 14 x + 49\right) = 109$

Simplifying this we get:
${x}^{2} + 7 x - 30 = 0$

This can be factorized to $\left(x - 3\right) \left(x + 10\right) = 0$ giving us the solutions $x = 3$ and $x = - 10$.

We can then solve for $y$ using the equation $y = x + 7$

If $x = 3$ then $y = 10$

If $x = - 10$ then $y = - 3$

Mar 28, 2018

$\left(x , y\right) = \left(3 , 10\right) \textcolor{w h i t e}{\text{xxx")"or"color(white)("xxx}} \left(x , y\right) = \left(- 10 , - 3\right)$

#### Explanation:

Given
[1]$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + {y}^{2} = 109$
[2]$\textcolor{w h i t e}{\text{XXX}} x - y = - 7$

Note that [2] implies
[3]$\textcolor{w h i t e}{\text{XXX}} x = y - 7$

Substituting $\left(y - 7\right)$ for $x$ in [1]
[4]$\textcolor{w h i t e}{\text{XXX}} {\left(y - 7\right)}^{2} + {y}^{2} = 109$

Expanding and simplifying the left side of [4]
[5]$\textcolor{w h i t e}{\text{XXX}} 2 {y}^{2} - 14 y + 49 = 109$

Converting to standard polynomial form by subtracting $109$ from both sides
[6]$\textcolor{w h i t e}{\text{XXX}} 2 {y}^{2} - 14 y - 60 = 0$

Factoring
[7]$\textcolor{w h i t e}{\text{XXX}} 2 \left(y - 10\right) \left(y + 3\right) = 0$

Which implies:
{: ([8a]color(white)("XX")y=10,color(white)("XX")"or"color(white)("XX"), [8b]color(white)("XX")y=-3), ("Substituting "10" for "x,,"Substituting "-3" for "x), ("in [3]",,"in [3]"), ([9a]color(white)("XX")x=10-7=3,,[9b]color(white)("XX")x=-3-7=-10) :}