Question

solve from substitution method
4x+y=24
3x-5y=18​

Answer 1

$Given \: system \: of \: equations :$

$4x + y = 24 \implies y = 24 - 4x \: --(1)$

$3x - 5y = 18 \: ---(2)$

/* Substitute y = 24 - 4x in equation (2) , we get */

$3x - 5(24-4x) = 18$

$\implies 3x - 120 + 20x = 18$

$\implies 23x = 18 + 120$

$\implies 23x = 138$

$\implies x = \frac{138}{23}$

$\implies x = 6$

/* Put x = 6 in equation (1), we get */

$y = 24 - 4 \times 6$

$\implies y = 24 - 24$

$\implies y = 0$

Therefore.,

$\green { x = 6 \: and \: y = 0 }$

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Answer 2

Givensystemofequations:

4x + y = 24 \implies y = 24 - 4x \: --(1)4x+y=24⟹y=24−4x−−(1)

3x - 5y = 18 \: ---(2)3x−5y=18−−−(2)

/* Substitute y = 24 - 4x in equation (2) , we get */

3x - 5(24-4x) = 183x−5(24−4x)=18

\implies 3x - 120 + 20x = 18⟹3x−120+20x=18

\implies 23x = 18 + 120⟹23x=18+120

\implies 23x = 138⟹23x=138

\implies x = \frac{138}{23}⟹x=

23

138

\implies x = 6⟹x=6

/* Put x = 6 in equation (1), we get */

y = 24 - 4 \times 6y=24−4×6

\implies y = 24 - 24⟹y=24−24

\implies y = 0⟹y=0

Therefore.,

\green { x = 6 \: and \: y = 0 }x=6andy=0