# How do you use substitution to solve the pair of equations: y=2x-3 and x^2+y^2=16?

##### 1 Answer
Dec 6, 2016

$\left(x , y\right) \approx \textcolor{g r e e n}{\text{(2.885,2770)) or color(green)(} \left(- 0.485 , - 3.970\right)}$

#### Explanation:

Given
[1]$\textcolor{w h i t e}{\text{XXX}} y = 2 x - 3$
[2]$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + {y}^{2} = 16$
Using [1] we know that we can substitute $\left(2 x - 3\right)$ for $y$ in [2]
[3]$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + {\left(2 x - 3\right)}^{2} = 16$

[4]$\textcolor{w h i t e}{\text{XXX}} \rightarrow {x}^{2} + 4 {x}^{2} - 12 x + 9 = 16$

[5]$\textcolor{w h i t e}{\text{XXX}} \rightarrow 5 {x}^{2} - 12 x - 7 = 0$

Using the quadratic formula:
[6]$\textcolor{w h i t e}{\text{XXX}} x = \frac{12 \pm \sqrt{{\left(- 12\right)}^{2} - 4 \cdot 5 \cdot \left(- 7\right)}}{2 \cdot 5}$

simplifying (and using a calculator)
color(white)("XXX"){:(x~~2.885," or ",x~~-0.485):}
using [1] $\rightarrow$
color(white)("XXX"){:(y~~2.770," or ", y~~-3.970):}

The image of the given relations (below) indicates that these results are reasonable.