Posts tagged with: systems of equations

Can you help me with Question #29 in Practice Test 3, Section 4? I found the answer by subtracting multiples of 9 from 122 to find one that was divisible by 5. Is there a better way?

 
The way I like to go here is to set up equations. If we say left-handed females are x and left-handed males are y, then we can fill in the table like this:

Now we have a system we can solve!

x+y=18
5x+9y=122

We eventually want a number of right-handed females, so let’s solve for x

5x+9(18-x)=122
5x+162-9x=122
40=4x
10=x

Remembering that x is the number of left-handed females, now we just need to multiply by 5 to get the right-handed females: 50.

The last trick here is that the question asks for a probability that a right-handed student being picked at random is female. There are 50 female right-handed students and 122 right-handed students overall, so the probability is \dfrac{50}{122}\approx 0.410.

Can you help with 33 in test 7 section 4?

Yep! This one is all about setting up the equations—translating words into math. The question tells us that the score in the game is calculated by subtracting the number of incorrect answers from twice the number of correct answers. Let’s say x is the number of correct answers and y is the number of incorrect answers. “Subtracting the number of incorrect answers” is easy enough: you’re going to have a ” – y” in your score equation. “Twice the number of correct answers” is pretty straightforward, too: if the number of correct answers is x, then twice that is 2x. So we can say that:

Score = 2xy

We’re not done though. We need to write another equation before we can solve. The question tells us the total number of questions answered; we need to use that information. The total number of questions answered must be the number of questions answered correctly plus the number answered incorrectly, right?

Questions = xy

Now we can plug in he values we know and solve the system of equations.

50 = 2xy
40 = xy

The question asks for the number the player answered correctly, so we need to solve for x. Conveniently, we can solve for x by eliminating y—all we need to do is add the equations together!

    50 = 2x – y
+ (40 = xy)
    90 = 3x

Of course, if 90 = 3x, then 30 = x. 30 is the answer.

Could you help me on number 9 on page 70 please? I kept getting a 4.312 answer

Sure. Here’s the problem:

The first thing you should do here is get rid of that infernal \sqrt{5}:

    \begin{align*}2x\sqrt{5}-3y\sqrt{45}&=12\sqrt{5}\\2x\sqrt{5}-3y\sqrt{9}\sqrt{5}&=12\sqrt{5}\\2x\sqrt{5}-3y(3)\sqrt{5}&=12\sqrt{5}\\2x\sqrt{5}-9y\sqrt{5}&=12\sqrt{5}\\2x-9y&=12\end{align*}

From there, you can subtract to eliminate the 9y terms:

2x – 9y = 12
– (5x – 9y = 18)
–3x = –6
= 2

Now use x = 2 to find y:

5(2) – 9y = 18
10 – 9y = 18
–9y = 8
y = –8/9

The question asks for the sum of x and y, so add them up!

2+\left(-\dfrac{8}{9}\right)=\dfrac{10}{9}

Could you help me number 10 on page 70 please? I kept getting 0 as the answer

Yep! Here’s the question:

I think the easiest way to go here is to use the calculator to graph. Don’t worry about the ≥ signs, just graph the lines and remember that the ≥ means anything on or above the lines you see.

Now, remember that you’re looking for the lowest possible y-coordinate that’s on or above both of those lines. In other words, you’re looking for the y-coordinate of the intersection! Since you’ve already graphed the lines on your calculator, it’s probably easiest just to use the calculator to find the intersection: (–1.33, 2.33).

2.33 is the answer you’d grid in (or its fraction form: \dfrac{7}{3}).

Solving algebraically is also not too cumbersome. Since both equations are in y = form already, substitute and solve for x:

    \begin{align*}-4x-3&=\dfrac{1}{2}x+3\\\\-6&=\dfrac{9}{2}x\\\\\-\dfrac{4}{3}&=x\end{align*}

Once you have x, plug it into either equation to get y:

    \begin{align*}y&=-4\left(-\dfrac{4}{3}\right)-3\\\\y&=\dfrac{16}{3}-\dfrac{9}{3}\\\\y&=\dfrac{7}{3}\end{align*}

 

Could you help me on number 4 of page 69 please? I forgot how to set up the equation

Sure. Here’s the question:

To write the equations, let’s say that c is the number of campers. The question gives us n for the number of lollipops.

“…if she were to give each camper 7 lollipops, she would have 10 left over”

That’s straightforward enough: if she would have 10 lollipops left over, then the number of lollipops must be 10 more than the product of 7 and the number of campers.

n = 7c + 10

“…if she eats one of the lollipops herself, she can give each camper 8 lollipops and have none left over”

This is a bit tricky because of her eating one, but the best way to think of that is that the number of lollipops she has is 1 greater than the product of 8 and the number of campers.

n = 8c + 1

Once we have the equations, all we need to do is solve for c. Let’s do that by elimination:

 n = 8c + 1
– (n = 7c + 10)
     0 = c – 9
     9 = c

Help on number 5 on page 69 please

Sure. The trick here is to look for opportunities to eliminate the variables you don’t want (i.e., y and z) by adding or subtracting. In this case, look what happens when you add all three equations together:

    \begin{align*}x+2y-3z&=92\\2x-y+z&=36\\4x-y+2z&=12\\\\7x+0y+0z&=140\\7x&=140\\x&=20\end{align*}

What’s the fastest way to do test 5 #18 in the no calculator section?

Because this is a no calculator question, it is worth your while to take a second to strategize before you dive in. In this case, a couple seconds of thinking will tell you two things:

  1. Those fractions will be easy to cancel out if you multiply by 2.
  2. You’re only asked to solve for x, so substitute accordingly.

First, cancel out the fraction:

    \begin{align*}\dfrac{1}{2}(2x+y)&=\dfrac{21}{2}\\2\left(\dfrac{1}{2}(2x+y)\right)&=2\left(\dfrac{21}{2}\right)\\2x+y&=21\end{align*}

Now take the second equation, y=2x, and substitute:

    \begin{align*}2x+y&=21\\2x+2x&=21\\4x&=21\\x&=\dfrac{21}{4}\end{align*}

You can grid in that fraction or its decimal equivalent: 5.25.

Test 3 Section 3 #9 please

When a system of linear equations has no solutions, that means the lines made by the equations in the system are parallel. Parallel lines have the same slope, so we should put both of these equations into slope-intercept form and then set up an equation to solve for k.

    \begin{align*}kx-3y&=4\\-3y&=-kx+4\\y&=\dfrac{k}{3}x-\dfrac{4}{3}\\ \\4x-5y&=7\\-5y&=-4x+7\\y&=\dfrac{4}{5}x-\dfrac{7}{5}\end{align*}

We see that one line has a slope of \dfrac{k}{3} and one has a slope of \dfrac{4}{5}. Remember, for the system to have no solutions, those slopes must be equal. So let’s solve for k.

    \begin{align*}\dfrac{k}{3}&=\dfrac{4}{5}\\k&=\dfrac{12}{5}\end{align*}

Can you explain number 9 on P. 72 of PWN the SAT 4th edition?

Sure. Recognize that \sqrt{45} simplifies to 3\sqrt{5}, so you can rewrite the first equation thusly: 2x\sqrt{5}-9y\sqrt{5}=12\sqrt{5}. Since there’s a \sqrt{5} in every term, you can just divide the whole equation by \sqrt{5} to eliminate the darn things! The equation simplifies to 2x-9y=12.

From there, this is a pretty easy problem. Solve for x by elimination:

    \begin{align*}2x-9y&=12\\-\left(5x-9y\right)&=18\\\rule{2cm}{0.4pt}&\rule{1cm}{0.4pt}\\-3x&=-6\\x&=2\end{align*}

From there, continue to solve for y:

    \begin{align*}5(2)-9y&=18\\10-9y&=18\\-9y&=8\\y&=-\dfrac{8}{9}\end{align*}

The question asks you for the sum of the coordinates of the ordered pair that satisfies the system of equations, so you have to add:

    \begin{align*}2+\left(-\dfrac{8}{9}\right)=\dfrac{10}{9}=1.\overline{1}\end{align*}

Gridding in either the fraction or the repeating decimal is fine as long as you fill all the boxes with the decimal (i.e., you enter “1.11”).

Can you explain number 7 on P. 72 of PWN the SAT 4th edition? Thank you.

Yep! Use elimination again here. Recognize that if you multiply the first equation by 5, you’ll have the easy opportunity to eliminate the x-terms.

    \begin{align*}5x\sqrt{2}+15y&=65\\-\left(5x\sqrt{2}+20y\right)&=-70\\\rule{2cm}{0.4pt}&\rule{1cm}{0.4pt}\\-5y&=-5\end{align*}

From there, you know that y = 1, so you know the answer must be A because no other choice has 1 for its y-coordinate. Might as well keep solving for x to be sure, though…

    \begin{align*}5x\sqrt{2}+15(1)&=65\\5x\sqrt{2}&=50\\x&=\dfrac{50}{5\sqrt{2}}\\x&=\dfrac{50}{5\sqrt{2}}\times\dfrac{\sqrt{2}}{\sqrt{2}}\\x&=\dfrac{50\sqrt{2}}{10}\\x&=5\sqrt{2}\end{align*}

Can you explain number 6 on P. 72 of PWN the SAT 4th edition? Thank you

Sure. For this question, you should see right away that the system lends itself to elimination of the y-terms. Subtract the second equation from the first:

    \begin{align*}3x-2y&=16\\-\left(\dfrac{2}{5}x-2y\right)&=-4\\\rule{2cm}{0.4pt}&\rule{1cm}{0.4pt}\\\dfrac{13}{5}x&=12\end{align*}

That tells you that x=\dfrac{60}{13}, which is enough to eliminate every choice but D.

You can also solve this by using your graphing calculator, as long as you get everything in = form first:

Now check the answer choices to see which line up with those decimals. Sure enough, choice D does.

Test 3 Section 3 #19

Let’s say h is the number of calories in each hamburger, and f is the number of calories in each order of fries. The question gives us enough information to write two equations:

hf + 50 (each hamburger has 50 more calories than each order of fries)
2h + 3f = 1700 (2 hamburgers and 3 orders of fries have a total of 1700 calories)

Let’s solve that system by substitution, since we already have h in terms of f.

2(f + 50) + 3f = 1700
2f + 100 + 3f = 1700
5f + 100 = 1700
5f = 1600
f = 320

The question asks us for h, so now we solve for h:

hf + 50
h = 320 + 50
h = 370