Hi Mike! Can you help me understand #26 on Math module 1, Digital SAT Test 1??

Sure thing. The most important thing to recognize quickly here is that the two equations we’re given, 2x+3y=7 and 10x+15y=35 are…the same! The second one is just the first one multiplied by 5. So any point that’s on one line is on both lines. So we can make life easy for ourselves and just work with the one with smaller numbers.

The question asks which of the points lies on the graph of 2x+3y=7 for all values of r, and at this point we’re going to have to figure out which of the choices makes the equation true. But if it’s gotta be true for all values of r, we can maybe save ourselves a little calculation by just testing an easy-to-work-with value of r. I see the choices dividing r by 2, 3, and 5, so let’s see what happens with the choices when we set r=30.

I don’t really like the look of choice A but let’s try it first. When r=30:

\left(\dfrac{30}{5}+7,-\dfrac{30}{5}+35\right)\\(13, 29)

Yeah, no way that’s going to satisfy our equation:

2x+3y=7\\2(13)+3(29)=7\\26+87\neq 7

Right, so choice A is no good, let’s do the same process with choice B, again setting r=30:

\left(-\dfrac{3(30)}{2}+\dfrac{7}{2},30\right)\\ \left(-45+3.5, 30\right)\\(-41.5,30)

Plug that into the equation:

2(-41.5)+3(30)=7\\-83+90=7\\7=7

Well that works great! At this point we’re feeling pretty good about choice B. We can either try to confirm choice B is good by plugging in the actual answer choice with the r-values, or I can see if r=30 eliminates the other choices.

I’ll go with the algebra this time.

The actual answer choice is \left(-\dfrac{3r}{2}+\dfrac{7}{2},r\right). Plug that into the equation:

2x+3y=7\\ 2\left(-\dfrac{3r}{2}+\dfrac{7}{2}\right)+3r=7\\\\-3r+7-3r=7\\7=7

Yeah, that works. So B is definitely the answer.

Does that help?

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