Question

Solve the following system of equations {y=1/6x+8 {y=3/4x+1 NEED HELP ASAP

Answer 1

x = 12 and y = 10

Solution:

Given system of equations are

$y=\frac{1}{6} x+8$ – – – – (1)

$y=\frac{3}{4} x+1$ – – – – (2)

Both are equations of y.

So, we can equate both the equations.

$\frac{1}{6} x+8=\frac{3}{4} x+1$

Arrange like terms one sides.

$\frac{1}{6} x-\frac{3}{4} x=1-8$

$\frac{1}{6} x-\frac{3}{4} x=-7$

Take LCM of the denominators.

LCM of 6 and 4 = 12

Make the denominators same.

$\frac{1\times 2}{6\times 2} x-\frac{3\times 3}{4\times 3} x=-7$

$\frac{2}{12} x-\frac{9}{12} x=-7$

$2x-9x=-7\times12$

$-7x=-84$

Divide both side of the equation by –7, we get

x = 12

Substitute x = 12 in equation (1).

$y=\frac{1}{6} \times 12+8$

$y=2+8$

y = 10

Hence x = 12 and y = 10.