Answer:
<em>2x+y=3 and 5x-y=11</em>
<em>AND</em>
<em>3x-2y=4 and -2x+2y=4</em>
both have only ONE solution.
Step-by-step explanation:
The possible answers are (given in the question):
2x+y=3 and 5x-y=11
2x+2y=1 and -2x-2y=1
3x+y=-1 and 6x+2y=-2
3x-2y=4 and -2x+2y=4.
Let's start with 2x+y=3 and 5x-y=11.
If you add both equations together, you get 7x=14, because the y's cancel out. This means x=2, but we still don't know what y is, so we can plug x=2 into the first equation. 2(2)+y=3. 2(2) is 4. 4+y=3. Subtract 4 on each side, you get y=-1.
The first system has only ONE solution, with x=2 and y=-1.
What about 2x+2y=1 and -2x-2y=1? This has NO solutions because when you add up the equations, it adds up to 0=2, because the 2x and -2x cancel out, and 2y and -2y also cancel out, which means on the left side you are left with 0, and 1+1 is 2 on the right side.
Next, 3x+y=-1 and 6x+2y=-2. This has an INFINITE number of solutions because 3x+y multiplied by 2 is 6x+2y and -1 times 2 is -2. These equations will form the same line, so anywhere on that line will be an correct answer for this equation.
Finally we come to 3x-2y=4 and -2x+2y=4.
When you add together these equations, it becomes x=4, because the -2y and 2y cancel out. Now plug in 4 in the first equation for x and you get 3(4)-2y=4. 3(4)=12, and now you have 12-2y=4. Subtract 12 on each side and you get -2y=8. Divide by -2 on each side and you are left with y=-4.
The last system has only ONE solution, with x as 4 and y as -4.
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