# How do you solve y+2x=5 and 2x^2 - 3x - y =16?

Mar 10, 2016

$A \left(- 3.11\right)$

$B \left(\frac{7}{2} , - 2\right)$

#### Explanation:

In this way:

$y + 2 x = 5$
$2 {x}^{2} - 3 x - y = 16$,

so:

$y = 5 - 2 x$

$2 {x}^{2} - 3 x - \left(5 - 2 x\right) = 16 \Rightarrow 2 {x}^{2} - x - 21 = 0$,

$\Delta = {b}^{2} - 4 a c = 1 - 4 \left(2\right) \left(- 21\right) = 1 + 168 = 169 = {13}^{2} \Rightarrow$

${x}_{1 , 2} = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{- \left(- 1\right) \pm 13}{2 \cdot 2} = \frac{1 \pm 13}{4} \Rightarrow$

${x}_{1} = \frac{1 - 13}{4} = - \frac{12}{4} = - 3$ and ${y}_{1} = 5 - 2 \cdot \left(- 3\right) = 11$
${x}_{2} = \frac{1 + 13}{4} = \frac{14}{4} = \frac{7}{2}$ and ${y}_{2} = 5 - 2 \cdot \frac{7}{2} = - 2$.

So:

$A \left(- 3.11\right)$

$B \left(\frac{7}{2} , - 2\right)$

Mar 10, 2016

$x = \frac{7}{2} \text{ or } x = - 3$

#### Explanation:

Given:
$y + 2 x = 5$........................(1)
$2 {x}^{2} - 3 x - y = 16$...........(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Write equation (1) as:$\text{ "y=-2x+5" } \ldots \ldots \ldots \ldots . . \left({1}_{a}\right)$
Write equation (2) as:$\text{ "y=2x^2-3x-16" } \ldots \ldots . \left({2}_{a}\right)$

Equate equation $\left({1}_{a}\right) \text{ to } \left({2}_{a}\right)$ through y

$\implies \text{ "-2x+5" "=" "y" "=" } 2 {x}^{2} - 3 x - 16$

Collecting like terms and equation to 0

$\text{ } 2 {x}^{2} - x - 21 = 0$

Standard form$\text{ } \to a {x}^{2} + b x + c = 0$

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

$a = 2$
$b = - 1$
$c = - 21$

Thus we have:

$\text{ } x = \frac{+ 1 \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(2\right) \left(- 21\right)}}{2 \left(2\right)}$

$\implies x = \frac{1 \pm \sqrt{169}}{4}$

$\implies x = \frac{1 \pm 13}{4}$

$\textcolor{b l u e}{\implies x = \frac{14}{4} = \frac{7}{2} \text{ or } x = - \frac{12}{4} = - 3}$ 