# How do solve the following linear system?:  4x+3y=1 , 2x +5y - 21 = 0 ?

Apr 15, 2016

$x = \frac{- 116}{28}$
$y = \frac{41}{7}$

#### Explanation:

Given -

$4 x + 3 y = 1$ ---------- (1)
$2 x + 5 y - 21 = 0$ -------- (2)

Take the constant term to the left in equation (1)

$4 x + 3 y - 1 = 0$

Let us use elimination method to solve the equations.

Multiply equation (2) with 2

$2 x + 5 y - 21 = 0$ -------- (2) $\times 2$
$4 x + 10 y - 42 = 0$

Equate both the equations

$4 x + 3 y - 1 = 4 x + 10 y - 42$

Take constant terms to right and other terms to left.

$4 x + 3 y - 4 x - 10 y = - 42 + 1$

Simplify -

$- 7 y = - 41$
$y = \frac{- 41}{- 7} = \frac{41}{7}$
$y = \frac{41}{7}$

Substitute $y = \frac{41}{7}$ in equation (1)

$4 x + 3 \left(\frac{41}{7}\right) = 1$
$4 x + \frac{123}{7} = 1$
$4 x = 1 - \frac{123}{7} = \frac{- 116}{7}$

$x = \frac{- 116}{7} \times \frac{1}{4} = \frac{- 116}{28}$
$x = \frac{- 116}{28}$