Help with really annoying systems of equations problem.

On one of my assignments, I'm asked to solve the following system of equations

-2x+3y=9

4x-6y=-18

The answer given for the system is (13/2,3/2). I've tried substitution and no matter how I arrange the numbers, I can't get that answer. Can someone please help me?

Re: Help with really annoying systems of equations problem.

There are an infinite number of solutions. The second equation is just -2 times the first equation. Choose a value for $x$, and then $y = 3+\dfrac{2}{3}x$.

The answer given doesn't even satisfy the given equations. Plugging into the first equation:

$-2\left(\dfrac{13}{2}\right) + 3\left(\dfrac{3}{2}\right) = -\dfrac{17}{2} \neq 9$.

Re: Help with really annoying systems of equations problem.

I hate Pearson textbooks. I'm just going to skip it. Thanks for trying to help me.

Re: Help with really annoying systems of equations problem.

Did you try this substitution: x= 13/2, y= 3/2? The first equation gives -2(13/2)+ 3(3/2)= -13+ 9/2= -26/2+ 9/2= -17/2, NOT 9. I don't know where you got that but x= 13/2, y= 3/2 is NOT a solution. You should have noticed immediately that -2(-2x+3y)= -2(9) is exactly the same as 4x- 6y= -18. ANY (x, y) pair that satisfies the first equation also satisfies the second and vice-versa.

For example, we can reduce -2x+ 3y= 9 to 3y= 9+ 2x and then y= 9/2- (2/3)x. In particular if x= 3/4, y= 9/2- (2/3)(3/4)= 9/2- 1/2= 8/2= 4 so x= 3/4, y= 4 satisfies both equations. But if x= 3/8, y= 9/2+ (2/3)(3/8)= 9/2+ 1/4= 18/4+ 1/4= 19/4 so that (3/8, 19/4) is another solution.

Re: Help with really annoying systems of equations problem.

Quote:

Originally Posted by

**HallsofIvy** For example, we can reduce -2x+ 3y= 9 to 3y= 9+ 2x and then y= 9/2- (2/3)x. In particular if x= 3/4, y= 9/2- (2/3)(3/4)= 9/2- 1/2= 8/2= 4 so x= 3/4, y= 4 satisfies both equations. But if x= 3/8, y= 9/2+ (2/3)(3/8)= 9/2+ 1/4= 18/4+ 1/4= 19/4 so that (3/8, 19/4) is another solution.

y = 3 + (2/3)x, not 9/2 - (2/3)x.