Question

Will mark brainliest if answered fast enough

Answer 1

Answer:

Option D that is y=-2(x+1) is the correct choice.

Explanation:

Given equation:

$y=2x-2$

We have to find another equation which has only one solution.

Standard form of equation:

$1.a_1x+b_1y+c_1=0$

$2.a_2x+b_2y+c_2=0$

Condition for one solution.

$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$

We have to arrange other equation in the form of  $ax+by+c=0$

Mark all the options as a,b,c and d.

$a.\ 2x+1-y=0$

$b.\ 2x-2-y=0$

$c.\ 2x-4-y=0$

$d.-2x-2-y=0$

Now from the above options we can see that option $a,b\ and\ c$ are not providing the desired result.

So

$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ can be obtained from the last option.

As

$\frac{a_1}{a_2} \neq \frac{b_1}{b_2} ,\frac{2}{-2} \neq \frac{-1}{-1} ,-1\neq 1$

Option D that is $y=-2(x+1)$ is the equation from where we can obtained only one solution.