Infinite number of solutions
Step-by step explanation:
Given system of equations are
[tex]y=\frac{1}{2} x+3[/tex] – – – – (1)
2y – x = 6 – – – – (2)
Equation (1) can be written as
⇒ [tex]y=\frac{1}{2} x+3[/tex]
Take LCM on both sides of the equation.
⇒ [tex]\frac{2}{2} y=\frac{1}{2} x+\frac{6}{2}[/tex]
⇒ 2y = x + 6
⇒ x – 2y + 6 = 0 – – – – (3)
Equation (2) can be written as
⇒ 2y – x = 6
⇒ x – 2y + 6 = 0 – – – – (4)
[tex]a_{1}=1, b_{1}=-2, c_{1}=6 \text { and } a_{2}=1, b_{2}=-2, c_{2}=6[/tex]
We know that the system of equations have infinite number of solutions if
[tex]\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}[/tex]
⇒ [tex]\frac{1}{1}=\frac{-2}{-2}=\frac{6}{6}[/tex]
Hence the given system have infinite number of solutions.
