Answer:
Option D that is y=-2(x+1) is the correct choice.
Explanation:
Given equation:
[tex]y=2x-2[/tex]
We have to find another equation which has only one solution.
Standard form of equation:
[tex]1.a_1x+b_1y+c_1=0[/tex]
[tex]2.a_2x+b_2y+c_2=0[/tex]
Condition for one solution.
[tex]\frac{a_1}{a_2} \neq \frac{b_1}{b_2}[/tex]
We have to arrange other equation in the form of [tex]ax+by+c=0[/tex]
Mark all the options as a,b,c and d.
[tex]a.\ 2x+1-y=0[/tex]
[tex]b.\ 2x-2-y=0[/tex]
[tex]c.\ 2x-4-y=0[/tex]
[tex]d.-2x-2-y=0[/tex]
Now from the above options we can see that option [tex]a,b\ and\ c[/tex] are not providing the desired result.
So
[tex]\frac{a_1}{a_2} \neq \frac{b_1}{b_2}[/tex] can be obtained from the last option.
As
[tex]\frac{a_1}{a_2} \neq \frac{b_1}{b_2} ,\frac{2}{-2} \neq \frac{-1}{-1} ,-1\neq 1[/tex]
Option D that is [tex]y=-2(x+1)[/tex] is the equation from where we can obtained only one solution.
