Question

Two linear equations are shown. A coordinate grid with 2 lines. The first line is labeled y equals StartFraction one-third EndFraction x plus 2 and passes through (negative 6, 0) and (0, 2). The second line is labeled y equals StartFraction 4 over 3 EndFraction minus 5. What is the solution to the system of equations? (7, 4) (7, StartFraction 13 over 3 EndFraction) (8, StartFraction 14 over 3 EndFraction) (9, 7)

Answer 1

Answer:

Its C.

{8, 14/3}

Answer 2

Answer:

$(7,\frac{13}{3})$

Step-by-step explanation:

we have

The equation of the first line

$y=\frac{1}{3}x+2$ ------> equation A

The equation of the second line

$y=\frac{4}{3}x-5$ ------> equation B

Solve the system of equations by elimination

Multiply equation A by -4 both sides

$(-4)y=(-4)(\frac{1}{3}x+2)$

$-4y=-\frac{4}{3}x-8$ --------> equation C

Adds equation B and equation C

$y=\frac{4}{3}x-5\\-4y=-\frac{4}{3}x-8\\--------\\y-4y=-5-8\\-3y=-13\\y=\frac{13}{3}$

Find the value of x

substitute the value of y

$\frac{13}{3}=\frac{1}{3}x+2$

$\frac{1}{3}x=\frac{13}{3}-2$

Multiply by 3 both sides

$x=13-6$

$x=7$

therefore

The solution to the system of equations is the point $(7,\frac{13}{3})$