Test 7 Section 4 Question 6

Hi Mike…SAT 7, Section 4, Q6: I now see the shortcut here (that both sides of the equation are perfect squares,) but if I did expand and FOIL the left side, wouldn’t I still get the correct “a” values even though it takes longer? I can’t get it to work !! Can you please show the alternate path math steps? Or is recognizing the perfect squares the ONLY way to solve this one ? Thanks!

SAT 8, Section 3, Number 7

Hi Mike, I’m asking about SAT 8, Section 3, Number 7. Is it always best to immediately plug in answer choices on questions like this? For the algebraic practice, I rewrote the equation as a quadratic and solved for (x=5) and (x= -1). Then sub’d each value back into the given equation to find that only (x=5) worked. OK…but what a time-killer at #7 out of 20. Any other solution path to consider? Thanks!

How would you use backsolving to solve practise test 2, section 4 question 29?

Hello, I know you’ve already solved practise test 2, section 4 question 29 (by either using your graphic calculator or by looking at the equation of a parabola) but how would you use backsolving? Lets say I try in option C and im getting y as 3 (which means my equations do NOT have 2 real solutions), how do I know whether to try out option B or D next?

Thank you so much!

For practise test 2, section 3 Q6, how exactly could I use backsolving to solve this?

Hi Mike! For practise test 2, section 3 Q6, how exactly could I use backsolving to solve this? Lets say I start with C and I plug in 8. My gradient of line l is 2/5. If I plug in p as 8, I’m getting gradient of line k as 4/8. Do I now compare the fractions? How do I know if I should try plugging in a bigger or smaller number to get closer towards 2/5 (initial gradient)?

Thanks!

Pwn the sat 4th edition page 36 question 3

Pwn the sat 4th edition page 36 question 3

I am tutoring my sophomore on her SAT math and we are going through your book a page at a time.

This problem is under back solving.
Given the amount of time you need to spend multiplying the left hand side of the equation. Wouldn’t it be faster to solve for a than plugging in the values?

How would you approach this differently?
Thanks