x = 12 and y = 10
Solution:
Given system of equations are
[tex]y=\frac{1}{6} x+8[/tex] – – – – (1)
[tex]y=\frac{3}{4} x+1[/tex] – – – – (2)
Both are equations of y.
So, we can equate both the equations.
[tex]$\frac{1}{6} x+8=\frac{3}{4} x+1[/tex]
Arrange like terms one sides.
[tex]$\frac{1}{6} x-\frac{3}{4} x=1-8[/tex]
[tex]$\frac{1}{6} x-\frac{3}{4} x=-7[/tex]
Take LCM of the denominators.
LCM of 6 and 4 = 12
Make the denominators same.
[tex]$\frac{1\times 2}{6\times 2} x-\frac{3\times 3}{4\times 3} x=-7[/tex]
[tex]$\frac{2}{12} x-\frac{9}{12} x=-7[/tex]
[tex]2x-9x=-7\times12[/tex]
[tex]-7x=-84[/tex]
Divide both side of the equation by –7, we get
x = 12
Substitute x = 12 in equation (1).
[tex]$y=\frac{1}{6} \times 12+8[/tex]
[tex]$y=2+8[/tex]
y = 10
Hence x = 12 and y = 10.
