# Solve the simultaneous equations 2x+y=8.......................................(1) 4x^2+3y^2=52.......................(2) ?

Dec 20, 2017

$x = 3.5 \mathmr{and} y = 1$
OR
$x = 2.5 \mathmr{and} y = 3$

#### Explanation:

$2 x + y = 8. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(1\right)$

$4 {x}^{2} + 3 {y}^{2} = 52. \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(2\right)$

(1) $\implies y = 8 - 2 x$

(2) $\implies 4 {x}^{2} + 3 {\left(8 - 2 x\right)}^{2} = 52$

$\implies 4 {x}^{2} + 3 \left(64 - 32 x + 4 {x}^{2}\right) = 52$

$\implies 4 {x}^{2} + 192 - 96 x + 12 {x}^{2} = 52$

$\implies 16 {x}^{2} - 96 x + 140 = 0$

$\implies 4 \left(4 {x}^{2} - 24 x + 35\right) = 0$

$\implies 4 {x}^{2} - 24 x + 35 = 0$
Solving this quadratic equation, we get:

$\implies \left(x - 3.5\right) \left(x - 2.5\right) = 0$

$\implies x = 3.5 \mathmr{and} x = 2.5$

Substitute this value of $x$ in equation (1):

Case 1: Taking $x = 3.5$

$\implies 2 x + y = 8$

$\implies 2 \left(3.5\right) + y = 8$

$\implies y = 8 - 7 = 1$

OR

Case 2 : Taking $x = 2.5$

$2 \left(2.5\right) + y = 8$

$\implies y = 8 - 5 = 3$

$\therefore x = 3.5 \mathmr{and} y = 1$

OR

$x = 2.5 \mathmr{and} y = 3$

Dec 20, 2017

$y = 3 \mathmr{and} x = \frac{5}{2} \mathmr{and} y = 1 \mathmr{and} x = \frac{7}{2}$

#### Explanation:

$2 x + y = 8$
$4 {x}^{2} + 3 {y}^{2} = 52$

$2 x = 8 - y \implies 4 {x}^{2} = {\left(8 - y\right)}^{2} = \textcolor{b l u e}{64 - 16 y + {y}^{2}}$

$\textcolor{b l u e}{64 - 16 y + {y}^{2}} + 3 {y}^{2} = 52$
$4 {y}^{2} - 16 y + 12 = 0$

dividing everything by 4(for symplifiying calculations):

${y}^{2} - 4 y + 3 = 0$

$y = \frac{4 \pm \sqrt{{4}^{2} - 4 \cdot 1 \cdot 3}}{2}$

$y = \frac{4 \pm \sqrt{4}}{2}$

$y = \frac{4 \pm 2}{2}$

$y = 3 \mathmr{and} x = \frac{5}{2} \mathmr{and} x = 1 \mathmr{and} x = \frac{7}{2}$