How do you solve the following system?: $x + 2 y = 6, 2 x - 3 y = 5$ $x + 2y = 6 , 2x - 3y = 5$

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Answer 1 · 2018-04-04T11:13:05

$x = 4 and y = 1$ $x=4 and y=1$

$x + 2 y = 6 (1)$ $x +2y =6" " " "(1)$
$2 x + 3 y = 5 (2)$ $2x + 3y =5" " " "(2)$

Multiply equation $(1)$ $(1)$ by $3$ $3$ and equation $(2)$ $(2)$ by $2$ $2$ ,

$3 \times (x + 2 y = 6)$ $3 xx (x +2y=6)$ 2 xx (2x -3y=5)#

Therefore,

$3 x + 6 y = 18 (3)$ $3x + 6y = 18" " " "(3)$
$4 x - 6 y = 10 (4)$ $4x - 6y = 10" " " "(4)$

Add equation $(3)$ $(3)$ to equation $(4)$ $(4)$

$7 x + 0 = 28$ $7x+0=28$

Make $x$ $x$ the subject of the equation

$x = \frac{28}{7}$ $x=28/7$

$x = 4$ $x=4$

Substitute the value of $x$ $x$ in equation $(1)$ $(1)$

$x + 2 y = 6$ $x +2y =6$

$4 + 2 y = 6$ $4+2y =6$

Collect like terms

$2 y = 6 - 4$ $2y= 6-4$

$2 y = 2$ $2y=2$

Divide through by $2$ $2$ to make $y$ $y$ the subject of the equation

$y = 1$ $y=1$

Ans:

$x = 4 and y = 1$ $x=4 and y=1$

How do you solve the following system?: x + 2y = 6 , 2x - 3y = 5 x+2y=6,2x−3y=5 x + 2y = 6 , 2x - 3y = 5