# How do you solve -x + y = 1 and x^2 +y^2 = 16 using substitution?

##### 2 Answers
Apr 29, 2017

$- x + y = 1$
$y = x + 1$ -----(1)

${x}^{2} + {y}^{2} = 16$ -----(2)

Substituting (1) into (2):
${x}^{2} + {\left(x + 1\right)}^{2} = 16$
${x}^{2} + {x}^{2} + 2 x + 1 = 16$
$2 {x}^{2} + 2 x - 15 = 0$

Solving the quadratic equation for x,
$x = \frac{\sqrt{31} - 1}{2} \mathmr{and} \frac{- \sqrt{31} - 1}{2}$

Substituting x into (1) to find y,
$y = \frac{\sqrt{31} + 1}{2} \mathmr{and} \frac{- \sqrt{31} + 1}{2}$

Apr 29, 2017

Bit long as I have explained a lot of the steps. See qin's solution for a shorter version (jumped steps)

$\left(2.28 , 3.28\right) \mathmr{and} \left(- 3.28 , - 2.28\right)$ to 2 dp

#### Explanation:

Recognise that ${x}^{2} + {y}^{2} = {4}^{2}$ is a Pythagorean (Pythagoras) representation of a circle

${x}^{2} + {y}^{2} = {4}^{2} \text{ } \ldots \ldots \ldots \ldots \ldots \ldots E q u a t i o n \left(1\right)$
$- x + y = 1 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(2\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{m a \ge n t a}{\text{Consider } E q u a t i o n \left(2\right)}$

Add $x$ to both sides giving:

$y = x + 1 \text{ } \ldots \ldots \ldots \ldots \ldots E q u a t i o n \left({2}_{a}\right)$

Using $E q u a t i o n \left({2}_{a}\right)$ substitute for $y$ in $E q u a t i o n \left(1\right)$

color(green)(x^2color(red)(+y^2)=4^2" "->" "x^2color(red)(+(x+1)^2)=16)" "..Equation(1_a)
.............................................................................................
$\textcolor{m a \ge n t a}{\text{Consider } {\left(x + 1\right)}^{2}}$

But ${\left(x + 1\right)}^{2} \to \textcolor{red}{\left(x + 1\right)} \textcolor{b l u e}{\left(x + 1\right)}$

Multiply everything inside the right bracket by everything inside the left bracket.

color(red)(xcolor(blue)((x+1))" "+" "1color(blue)((x+1))
$\textcolor{w h i t e}{.} {x}^{2} + x \text{ "color(white)(.)+" } x + 1$

${x}^{2} + 2 x + 1$
.....................................................................................
$\textcolor{m a \ge n t a}{\text{Putting it all together to determine } x}$

$E q u a t i o n \left({1}_{a}\right)$ becomes:

${x}^{2} + {x}^{2} + 2 x + 1 = 16$

$2 {x}^{2} + 2 x + 1 = 16 \text{ "->" } 2 {x}^{2} + 2 x - 15 = 0$

Using the standardised form $y = a {x}^{2} + b x + c$

where $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$a = 2 \text{; "b=2"; } c = - 15$

$\implies x = \frac{- 2 \pm \sqrt{{\left(2\right)}^{2} - 4 \left(2\right) \left(- 15\right)}}{2 \left(2\right)}$

$x = - \frac{1}{2} \pm \frac{\sqrt{124}}{4}$

$x = - \frac{1}{2} \pm \frac{\sqrt{{2}^{2} \times 31}}{4} \text{ "->" } x = - \frac{1}{2} \pm \frac{1}{2} \sqrt{31}$

$x \approx 2.28388 \ldots \mathmr{and} x \approx - 3.28388 \ldots$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{m a \ge n t a}{\text{Determine "y" by substitution for "x" in } E q u a t i o n \left({2}_{a}\right)}$

$y = x + 1 \text{ } \ldots \ldots \ldots \ldots . . E q u a t i o n \left({2}_{a}\right)$

Set $x \approx 2.28388 \ldots$

$y \approx 3.28288 \ldots$

$\underline{\overline{| \textcolor{w h i t e}{\frac{2}{2}} \text{point "P_1->(x,y)=(2.28,3.28)" to 2 decimal places }} |}$

......................................................................................

Set $x \approx - 3.28388 \ldots$

$y \approx - 2.28288 \ldots$

ul(bar(|color(white)(2/2)"point "P_2->(x,y)=(-3.28,-2.28)" to 2 decimal places "