Question

Please help me solve question 2

Answer 1

Answer:

solution: ( 3 , 6 )

$\sf y =6$

$\sf x =3$

Step-by-step explanation:

$\sf y = \frac{2}{3} x+4$

$\sf 2x+3y = 24$

make y the subject in equation 2:

$\sf 2x+3y = 24$

$\sf 3y = 24-2x$

$\sf y = \frac{24-2x}{3}$

insert this in equation 1:

$\sf \frac{24-2x}{3} =\frac{2}{3} x+4$

$\sf \sf \frac{24-2x}{3} =\frac{2x+12}{3}$

$\sf (24-2x) = (2x+12)$

$\sf -2x-2x=12-24$

$\sf -4x=-12$

$\sf x =3$

solve for y:

$\sf y = \frac{24-2x}{3}$

$\sf y = \frac{24-2(3)}{3}$

$\sf y =6$

Answer 2

Answer:

(3, 6)

Step-by-step explanation:

$\textsf{Equation 1}: \ y=\dfrac23x+4$

$\textsf{Equation 2}: \ 2x + 3y =24$

Substitute equation 1 into equation 2:

$\implies 2x + 3\left(\dfrac23x + 4 \right) =24$

Expand brackets:

$\implies 2x +2x+12=24$

Collect like terms:

$\implies 4x+12=24$

Subtract 12 from both sides:

$\implies 4x=12$

Divide both sides by 4:

$\implies x=3$

Now substitute found value of x into equation 1:

$\implies y=\dfrac23(3)+4$

$\implies y=2+4$

$\implies y=6$

Therefore, solution is (3, 6)