How do you solve the following system: $2 x + y = 8, 3 x - 2 y = 24$ $2x + y = 8 , 3x -2y=24$ ?

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Answer 1 · 2018-05-09T03:52:23

$x = \frac{40}{7}, y = - \frac{24}{7}$ $color(maroon)(x = 40/7, y = -24/7$

$2 x + y = 8, Eqn (1)$ $2x + y = 8, " Eqn (1)"$

$3 x - 2 y = 24, Eqn (2)$ $3x - 2y = 24, " Eqn (2)"$

$4 x + 2 y = 16, Eqn(1) * 2$ $4x+ 2y = 16, " Eqn(1) * 2"$

$Adding Eqn(1) * 3 + Eqn(2)$ $"Adding Eqn(1) * 3 + Eqn(2)"$

$7 x = 40$ $7x = 40$

$x = \frac{40}{7}$ $color(maroon)(x = 40/7$

Substituting value of x in Eqn (1),

$\frac{2 \cdot 40}{7} + y = 8$ $(2*40)/7 + y = 8$

$y = 8 - \frac{80}{7} = - \frac{24}{7}$ $color(maroon)(y = 8 - 80/7 = -24/7$

How do you solve the following system: 2x+y=8,3x−2y=242x + y = 8 , 3x -2y=24 ?