The three Math Idol judges have been eliminating contestants all day! The number of one-step equations and two-step equations who have been eliminated today is equal to 1150. If three times the number of one-step equations minus twice the number of two-step equations is equal to 1300, how many one-step equations auditioned today?

430

720

1150

1300

Question

The three Math Idol judges have been eliminating contestants all day! The number of one-step equations and two-step equations who have been eliminated today is equal to 1150. If three times the number of one-step equations minus twice the number of two-step equations is equal to 1300, how many one-step equations auditioned today?

 430

 720

 1150

 1300

shane_b · Answer

Let x be the number of one-step equations and y be the number of two-step equations:
x+y=1150
3x-2y=1300

Can you solve it from here?

anonymous · Answer

no i dont get it.

shane_b · Answer

Do you at least understand how the equations above were generated?  If so, I'll show you the rest.

anonymous · Answer

you put it all together from the word problem

shane_b · Answer

Yes, if you understand how I did that, we can move on to solve the system of equations.  That part is pretty easy in this problem.

shane_b · Answer

Ok...moving on.  We developed two equations which have two unknowns:
x+y=1150 
3x-2y=1300

Our goal is to solve for x and y.  I believe the simplest way to do that is to solve one of the equations for a single variable and then substitute that back into the other equation.  

Looking at the first equation, let's rearrange the terms:
$$x+y=1150 ==>x=1150-y$$
Now take the second equation and plug in "1150-y" anywhere you see an x:
$$3x-2y=1300==>3(1150-y)-2y=1300$$With me so far?

shane_b · Answer

Guess not :)  If I'm here when you get back we can finish this...or someone else can.

anonymous · Answer

i would write the same answer with @Shane_B