Posts tagged with: backsolve

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!

FOIL should work, sure. Plug in –3 for x and go to town:

    \begin{align*}(-3a+3)^2&=36\\9a^2-18a+9&=36\\a^2-2a+1&=4\\a^2-2a-3&=0\\(a-3)(a+1)&=0\end{align*}

That tells you that a could equal –1 or 3, and –1 is an answer choice.

You can also get this one by backsolving, though. That would actually be my first choice. You’re given x = –3. Try the answer choices for a until one works! Conveniently, I usually start with C when I backsolve, so I finish this one very quickly:

Choice C:

((-1)(-3)+3)^2=36

6^2=36

36=36 <– checks out

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!

For questions like this (extraneous solution quadratics) the fastest solution IS to backsolve from the answer choices. If you look at the choices right away, you see that the only possible answers are 0, –1, and 5. Check those three out and you’re done in no time. Solve algebraically and you have to check the answers you get anyway to make sure they’re not extraneous!

To drive it home, here’s all the work you need to do if you backsolve:

    \begin{align*}\sqrt{2(0)+6}+4&=0+3\\\sqrt{6}+4&=3\end{align*}

    \begin{align*}\sqrt{2(-1)+6}+4&=-1+3\\\sqrt{4}+4&=2\\2+4&=2\\6&=2\end{align*}

    \begin{align*}\sqrt{2(5)+6}+4&=5+3\\\sqrt{16}+4&=8\\4+4&=8\\8&=8\end{align*}

Obviously only the last one of those worked out correctly, so the solution set contains only 5.

If you solve algebraically, as you point out, you eventually end up at –1 and 5 as possible answers (work below for anyone still reading), but because you had to square everything to get there, you need to check for extraneous solutions by doing the step you could have started with if you just backsolved from the beginning!

    \begin{align*}\sqrt{2x+6}+4&=x+3\\\sqrt{2x+6}&=x-1\\2x+6&=x^2-2x+1\\0&=x^2-4x-5\\0&=(x-5)(x+1)\end{align*}

 

College Board Test 4 Section 3 #9:
____
√x-a = x-4
If a=2 what is the solution set of the preceding equation?
A. {3, 6}
B. {2}
C. {3}
D. {6}

Is there another way to solve this quickly besides plugging in the answer choices?

Backsolve is far and away my preferred method on a question like this. The answer choices only show you 2, 3, and 6 as possible solutions, so you have to try, at most, three numbers.

However, of course there is an algebraic solution! If you go that way, you must be careful because you will generate extraneous solutions. To begin, square both sides, combine like terms, and solve the resulting quadratic:

    \begin{align*}\sqrt{x-2}&=x-4\\\left(\sqrt{x-2}\right)^2&=(x-4)^2\\x-2&=x^2-8x+16\\0&=x^2-9x+18\\0&=(x-6)(x-3)\end{align*}

From there, you must check that both solutions you generated actually work in the original equation, because when you square both sides of an equation you can cause extraneous solutions.

    \begin{align*}\sqrt{3-2}&=3-4\\1&\ne -1\end{align*}

    \begin{align*}\sqrt{6-2}&=6-4\\2&=2\end{align*}

As you can see, 3 doesn’t work in the original equation, so it’s an extraneous solution. 6 DOES work in the original equation, so it’s the one we go with.

Hopefully this reinforces the appeal of backsolving here: even if you do the algebra, you still need to try the answers you find in the original equation anyway. Why not just start by trying the answers provided!? 🙂

Test 8 Section 4 #8

This is a good one to backsolve! Note that you’re asked for a value of x + 1 (as opposed to a value for x) so backsolving is really easy. Start with the easiest ones to try.

    \begin{align*}x+1&=\dfrac{2}{x+1}\\\\\text{C)     }\ \ \qquad 2&=\dfrac{2}{2}\\\\2&\ne1\\\\\text{D)     }\ \ \qquad 4&=\dfrac{2}{4}\\\\4&\ne 0.5\\\\\text{B)     }\qquad\sqrt{2}&=\dfrac{2}{\sqrt{2}}\\\\\sqrt{2}&=\dfrac{2}{\sqrt{2}}\times\dfrac{\sqrt{2}}{\sqrt{2}}\\\\\sqrt{2}&=\dfrac{2\sqrt{2}}{2}\\\\\sqrt{2}&=\sqrt{2}\end{align*}

Choice B looks good to me!

Test 5 Section 4 Number 23

All we really need to do to nail this question is match a table to a linear function. Unlike some similar questions, the table isn’t labeled with variables and the values aren’t sorted in ascending or descending order, but those are just details! We don’t need to get too creative here; let’s just backsolve through the choices and see which one works best. Remember that r represents the monthly rental price (the rightmost column) and p represents the purchase price in thousands (the middle column).

Let’s drop the first row of values (Clearwater Lane) into choice A:

    \begin{align*}r(p)=2.5p-870\\950=2.5(128)-870\\950=-550\end{align*}

NOPE! That didn’t work. We can eliminate A. Let’s try B:

    \begin{align*}r(p)=5p+165\\950=5(128)+165\\950=805\end{align*}

That’s not quite as ridiculous, but still isn’t close enough. Let’s try C:

    \begin{align*}r(p)=6.5p+440\\950=6.5(128)+440\\950=1275\end{align*}

No way, that’s not even close. Eliminate C. Try D:

    \begin{align*}r(p)=7.5p-10\\950=7.5(128)-10\\950=950\end{align*}

Huh. That’s pretty good! To make sure D is the answer, let’s try a couple other rows of data: Driftwood Drive and Edgemont Street.

    \begin{align*}r(p)=7.5p-10\\1310=7.5(176)-10\\1310=1310\end{align*}

    \begin{align*}r(p)=7.5p-10\\515=7.5(70)-10\\515=515\end{align*}

Both are perfect! That’s enough to satisfy me. D is the answer.

Test 3 Section 4 #24

Of course, we could use algebra to solve this (and I will down below), but because the answer choices are numbers that can easily be dropped back into the word problem, my recommended strategy is backsolving.

Start by trying C; assume 23 students in the class. If Mr. Kohl gives each student 3 ml, he’ll have 5 ml left over. 3(23)+5=74. For Mr. Kohl to give each student 4 ml, he’ll need an additional 21 ml. 4(23)=92. Is that 21 higher than 74? It is not—it’s only 18 more.

That suggests we need even more kids in the class, so the answer is almost certainly D, which says there are 26 students. If Mr. Kohl gives each student 3 ml, he’ll have 5 ml left over. 3(26)+5=83. For Mr. Kohl to give each student 4 ml, he’ll need an additional 21 ml. 4(26)=104. Is that 21 higher than 83? Yes! D is the answer.

Now, the algebra. What we want to do here is come up with two expressions that are both equal to n, the amount of solution Mr. Kohl has. I’ll use x for the number of students he has.

If Mr. Kohl gives each student 3 ml, he’ll have 5 ml left over. n=3x+5

In order to give each student 4 ml, he will need an additional 21 ml. n=4x-21

Now that we have those equations, we can solve by substitution.

    \begin{align*}3x+5&=4x-21\\5&=x-21\\26&=x\end{align*}

Test 7 Section 3 #15

If you recognize that \dfrac{1}{3}(x+k)(x-k) contains a factored difference of two squares, then you can work from there to get this one very quickly.

    \begin{align*}&\dfrac{1}{3}(x+k)(x-k)\\=&\dfrac{1}{3}\left(x^2-k^2\right)\\=&\dfrac{1}{3}x^2-\dfrac{1}{3}k^2\end{align*}

If you know that has to equal \dfrac{1}{3}x^2-2, then you know that \dfrac{1}{3}k^2=2. Which answer choice makes that so? Yep, k=\sqrt{6} does:

    \begin{align*}&\dfrac{1}{3}x^2-\dfrac{1}{3}k^2\\=&\dfrac{1}{3}x^2-\dfrac{1}{3}\left(\sqrt{6}\right)^2\\=&\dfrac{1}{3}x^2-\dfrac{1}{3}(6)\\=&\dfrac{1}{3}x^2-2\end{align*}

 

Could you help me with question 9 on page 38 please? I don’t understand how to solve it even using back-solving

Yeah, this one is pretty tough! 🙂

To backsolve it successfully, it’s helpful to also plug in. Say that the base and height of triangle A are both 10. Then if you try, say, answer choice C, which says that p = 23, you know that r = 18, and therefore that the base of triangle B is 12.3 (23% more than 10) and the height is 8.2 (18% less than 10). Does that give you the same area for both triangles?

\text{Area}_A=\dfrac{1}{2}(10)(10)=50

\text{Area}_B=\dfrac{1}{2}(12.3)(8.2)=50.43

Not quite, but you’re close! For that reason, probably makes sense to try the next closest choice, D. Sure enough, that one works.

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!

You wouldn’t really know whether to try B or D next without knowing that you’re playing with whether the parabola opens up or down and its y-intercept when you change a and b, respectively. This is important, though: don’t waste more than a couple seconds trying to figure out which way to go! If you don’t know which choice to try next right away, pick any one. If you’re backsolving simply by changing a graph on your calculator, it shouldn’t take more than a couple seconds to try each choice until one works.

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!

Right, you’re comparing the fractions. (Of course, this requires you to know that parallel lines have the same slope.)

To decide which direction to go, recognize that \dfrac{4}{8}=\dfrac{1}{2} is bigger than \dfrac{2}{5}. You need your fraction to get smaller. Since the value you’re playing with is p, which is in the denominator, you make the fraction smaller by increasing the denominator.

A couple notes:

  1. Generally speaking, don’t spend too much time trying to figure out which answer choice to try next if it’s not obvious. It’s probably faster just to guess at which direction to go and see if you get closer or farther away from where you want to be!
  2. Just because I say a problem CAN be solved by backsolving doesn’t mean I’m saying that backsolving is the best way for that problem. I just want you to train yourself to recognize how often it’s possible to backsolve so that when you encounter a problem that can be backsolved on test day, you’re not blind to the possibility.

Hi. Can you explain test 6, section 4, number 14. Thanks.

Sure. The table tells you that the sunflower’s height was 36.36 cm on day 14 and 131.00 cm on day 35. The equation that models that interval well should give you about 36 cm when you substitute 14 for t and give you about 131 cm when you substitue 35 for t. Therefore, you can try each choice using those numbers until you find the one that works.

A) h = 2.1(14) – 15 = 14.4 <– Nope, too small!

B) h = 4.5(14) – 27 = 36 <– Looks good, better check the other value.
h = 4.5(35) – 27 = 130.5 <– Oh yeah, that’s pretty darn close. That’s gotta be our answer.

The other way to go here is to calculate the slope between the two points you know:

\text{slope}=\dfrac{h_2-h_1}{t_2-t_1}=\dfrac{131.00-36.36}{35-14}=4.50666...

The only answer choice with a slope of 4.5 is B, so again, that must be the answer!

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

The question:

I get the “Wouldn’t it be faster…” question a lot about both the plug in and backsolve techniques, and my stock answer is maybe. In this case, I agree with you that FOILing the left hand side until you get the right answer (as many as 3 times—if you fail 3 times then you know the 4th choice is correct) might take a minute. I also agree that if you know what you’re doing with complex numbers, solving for a is fairly straightforward. To make a stronger case for backsolving here, I maybe should have made the question a tiny bit tougher. Like this, perhaps:

(9+ai)(1-i)=12-2ai

As I say at the top of that drill, though, I think it’s worth practicing the techniques, even if they feel slower or actually are slower, because if you really internalize the technique, it can be an escape hatch for you on test day when you encounter a question you can’t figure out the math way. If you never practice backsolving because it always feels too slow, then it probably won’t be there for you in a pinch when you need it most.

Here’s how backsolving would look for me on this question. I’d start with C, since I pretty much always start with C:

    \begin{align*}(9+2i)(1-i)&=12-6i\\9-9i+2i-2i^2&=12-6i\\9-7i+2&=12-6i\\11-7i&=12-6i\end{align*}

That’s obviously false, but it’s CLOSE! Being 1 off would almost certainly clue me in to trying choice B next, and I’d follow the pattern of the FOILing I just did, so I’d be a little faster the 2nd time.

    \begin{align*}(9+3i)(1-i)&=12-6i\\9-9i+3i-3i^2&=12-6i\\9-6i+3&=12-6i\\12-6i&=12-6i\end{align*}

 

Hi! Can you please show me how problem 8 is solved in test 2 section 3?

Sure! First, don’t get bogged down in the wall of text. You should read it, of course, but you should read it while remembering that the formula you’re going to need has already been given to you at the top of the question. A simplified version of the question is this:

nA = 360. If A > 50, what’s the greatest integer value of n?

I’ll suggest two ways to solve. First, you can backsolve. The question asks for the greatest number of sides, so start with the largest answer choice, 8, and see if you get a value of A that’s greater than 50 when you set n = 8.

8A = 360
A = 45

Nope, not quite. So try n = 7.

7A = 360
A = 51.4…

That works! So choice C is correct.

The mathier way is to use an inequality. First, get A by itself in the given equation:

A=\dfrac{360}{n}

Now, remembering that we know A must be greater than 50 and therefore \dfrac{360}{n} must be greater than 50, make an inequality.

50<\dfrac{360}{n}

50n<360

n<7.2

Since the question asked for the largest possible value of n, which can’t be a decimal because n represents a number of sides and a polygon can’t have a decimal number of sides, we go with 7.

Will you please answer question # 27, Test 3, section 4. Is it possible to use real numbers in your example? Thanks!

Sure! The thing you want to remember when you’re plugging in numbers is that you’re doing it to make your life easier, so pick numbers that will really make your life easier. In this case, that means I’m going to say that the original rectangle is a 10 by 10 square, so its area is 100. Obviously, it’s very easy to work with 100 when dealing with percents, and it’s also easy to take different percents of 10 to increase/decrease side lengths.

So, yeah. Here’s our original rectangle:

Now we need to increase its length by 10 percent (10 percent of 10 is 1, which means the length increases from 10 to 11), and decrease its width by p percent. What should we do for p? Well, why not backsolve?

Say p = 20, like answer choice C says. (I’m picking that because it’s in the middle, and also because 20 is the easiest choice to work with so why not try it first.) If p = 20, then we decrease the width by 2, taking it from 10 to 8. So here’s the new rectangle:

What’s the difference between the original area of 100 and the new area of 88? 12. And here’s where we pat ourselves on the back for picking numbers that gave us 100 as our starting area: 12 is what percent of 100? 12 percent! So the first choice we tried, C, is right. Awesome!

(Whenever I illustrate backsolving and the first choice I try is the right choice, I feel compelled to show what a wrong choice would look like. So let’s quickly look at choice B. We’re still increasing the length by 10 percent, so the length is still 11. The width is now decreased by 15 percent, making it 8.5. The new area is 11\times 8.5=93.5. Is that a 12 percent decrease from the original area of 100? No, 93.5 is too big! So if we didn’t already know the answer, we’d know that the next choice we should try should be smaller.)

Test 4 Section 3 #13

You’re told that the given parabolas intersect at (k,0) and (-k,0). Important insight: when two graphs intersect at a point, that point is on both graphs. Therefore, we don’t need to use both equations to solve this question, we can just pick one and find the values of x that will result in a function value of zero.

f(x)=8x^2-2

f(k)=8k^2-2

0=8k^2-2

2=8k^2

\dfrac{1}{4}=k^2

\pm\dfrac{1}{2}=k

The given figure tells you that k is positive (look where k and -k are on the x-axis) so you know the answer is \dfrac{1}{2}.

Note: You can also backsolve this one. Put the answer choices into one of the functions until you get a result of 0. Even without a calculator, that won’t take long (but will probably take longer than the algebra unless you’re stuck).