Respuesta :
Answer:
(- [tex]\frac{4}{3}[/tex], 4 )
Step-by-step explanation:
Given the 2 equations
3x + 8y = 28 → (1)
3x + 2y = 4 → (2)
Subtracting (2) from (1) term by term will eliminate the x- term
(3x - 3x) + (8y - 2y) = (28 - 4), that is
6y = 24 ( divide both sides by 6 )
y = 4
Substitute y = 4 into either of the 2 equations and solve for x
Substituting into (1)
3x + 8(4) = 28
3x + 32 = 28 ( subtract 32 from both sides )
3x = - 4 ( divide both sides by 3 )
x = - [tex]\frac{4}{3}[/tex]
Solution is (- [tex]\frac{4}{3}[/tex], 4 )
If you subtract the two equations you get
[tex]\underbrace{(3x + 8y)-(3x + 2y)}_{\text{Difference of the LHS}}=\underbrace{28-4}_{\text{Difference of the RHS}}[/tex]
This can be simplified into
[tex]6y=24 \iff y=4[/tex]
Now you can plug this value for [tex]y[/tex] in one of the two equations, and solve it for [tex]x[/tex].
