Hot X: Algebra Exposed

Solution Guide for Chapter 12 Here are the solutions for the “Doing the Math” exercises in Hot X: Algebra Exposed!

DTM from p.168

2. Let’s plug in 3x for y in the second equation, and we get:

y – x = 4 3x – x = 4 2x = 4 x = 2.

Great! We have one of our values. Now we can stick in 2 for x into either original

equation (how about y = 3x) and solve for y:

y = 3x y = 3(2) y = 6.

To check our answer, let’s plug x = 2 and y = 6 into the other original equation (y – x = 4)

to see if we get a true statement:

y – x = 4 6 – 2 = 4? 4 = 4 Yep!

Answer: x = 2, y = 6

3. Notice that the first equation can be simplified: We’ll multiply both sides by 12

and

get: 2y + 8x = 2 122y + 8x( ) = 1

2(2) y + 4x = 1

This is definitely the simpler of the two equations, and we’ll subtract 4x from both sides

to use it for substitution: y = 1 – 4x. Now we can plug in 1 – 4x for y in the second

equation to solve for x:

2y + 7x = 3 2(1 – 4x) + 7x = 3 2 – 8x + 7x = 3 2 – 1x = 3

–x = 1 x = –1

Great! Now we can plug this value into either original equation, how about 2y + 8x = 2,

and we get: 2y + 8x = 2 2y + 8(–1) = 2 2y – 8 = 2 2y = 10

y = 5.

To check our answer, we’ll plug these values into the other original equation,

2y + 7x = 3, and see if we get a true statement:

2y + 7x = 3 2(5) + 7(–1) = 3 ? 10 – 7 = 3? 3 = 3? Yep, we sure did!

Answer: x = –1, y = 5

4. Let’s go ahead and multiply both sides of the first equation by 8, to get rid of that

fraction, and we get 8y = x. Cool, now we can substitute 8y where we see x in the other

equation:

2x – 4y = 3 2(8y) – 4y = 3 16y – 4y = 3 12y = 3 y = 14

Okay, now we’ll plug this into one of the equations – how about 2x – 4y = 3, and we get:

2x – 4y = 3 2x − 414

⎛⎝⎜

⎞⎠⎟= 3 2x – 1 = 3 2x = 4 x = 2

And let’s test these values, x = 2 and y = 14

, in the other original equation to see if we

found the right values: y = x8

14= 28

? 14=14

Yep! Done!

Answer: x = 2, y = 14

DTM from p.174

2. I see some evil twins: 3y and –3y! That means we can just add these two equations

together in order to eliminate the y variable:

5x – 3y = 19

2x + 3y = –5

+_____________

7x + 0 = 14

And we can easily solve that: 7x = 14 x = 2. Great! Let’s plug this into the first

equation and solve for y:

5x – 3y = 19 5(2) – 3y = 19 10 – 3y = 19 –3y = 9 y = –3.

To check our answer, let’s plug in x = 2 and y = –3 into the other original equation and

see if we get a true statement:

2x + 3y = –5 2(2) + 3(–3) = –5? 4 – 9 = –5? –5 = –5? Yep!

Answer: x = 2, y = –3

3. This time I see twins! 3x and 3x. So we can subtract these two equations in order to

eliminate the x variable:

3x – 5y = –36 scratch work:

3x + 5y = 12 –5y – 5y = –10y

– _________________ –36 – 12 = –48

0 – 10y = –48

And we can solve this by dividing both sides by –10, and reducing:

–10y = –48 y =−48−10

y = 245

Okay, now we can plug this into either equation, how about 3x + 5y = 12, and get:

3x + 5y = 12 3x + 5 245

⎛⎝⎜

⎞⎠⎟= 12 3x + 24 = 12 3x = –12 x = –4.

And now we’ll plug both our new values into the other original equation, 3x – 5y = –36,

and get: 3x – 5y = –36 3(–4) − 5 245

⎛⎝⎜

⎞⎠⎟= −36 ? –12 – 24 = –36?

–36 = –36? Yep!

Answer: x = –4, y = 245

4. The QUICK NOTE on p.172 reminds us that sometimes we need to multiply both

sides of BOTH equations in order to get twins. Hm, y’s coeffcients have more factors in

common – we can multiply them times small numbers in order to get “30y twins” so let’s

multiply the first equation by 3 and the second equation by 2, and we get:

7x + 10y = 17 3(7x + 10y) = 3(17) 21x + 30y = 51

and: 8x + 15y = 23 2(8x + 15y) = 2(23) 16x + 30y = 46

Great, now we have “30y” twins! And we’re ready to subtract these equations:

21x + 30y = 51 scratch work:

16x + 30y = 46 21x – 16x = 5x

–_____________ 51 – 46 = 5

5x + 0 = 5

Which is easy to solve: 5x = 5 x = 1! Now we can plug x = 1 into one of the original

equations to solve for y; how about the first one:

7x + 10y = 17 7(1) + 10y = 17 7 + 10y = 17 10y = 10 y = 1.

Great, now let’s check our work by testing these values into the second original equation:

8x + 15y = 23 8(1) + 15(1) = 23 ? 8 + 15 = 23 ? 23 = 23 Yep!

Answer: x = 1, y = 1

5. Hm, again we’ll need to multiply both sides of both equations by stuff in order to get

twins. Let’s make “6y” twins; we’ll multiply the first equation by 2 and the second

equation by 3, and we get:

4x + 3y = –7 2(4x + 3y) = 2(–7) 8x + 6y = –14

and 3x + 2y = –6 3(3x + 2y) = 3(–6) 9x + 6y = –18

Now we have “6y” twins, so we can subtract the two equations and eliminate the y

variable:

8x + 6y = –14 scratch work:

9x + 6y = –18 8x – 9x = –1x

– ______________ –14 – (–18) = 4

–1x + 0 = 4

And we solve for x pretty easily: –1x = 4 x = –4. Great! Now we’re ready to plug

this value of x into one of the original equations to find y. Let’s use 4x + 3y = –7, and we

get:

4x + 3y = –7 4(–4) + 3y = –7 –16 + 3y = –7 3y = 9 y = 3.

Okay, now we’ll take our two values, x = –4 and y = 3, and test them in the other original

equation: 3x + 2y = –6 3(–4) + 2(3) = –6 ? –12 + 6 = –6? –6 = –6?

Yep!

Answer: x = –4, y = 3

DTM from p.180-181

2. These lines, both in slope-intercept form, have two different slopes. That means they

must intersect somewhere, and that means they have one (x, y) solution – in other words,

the two lines are consistent & independent. Time for part b, where we find out what this

(x, y) solution is!

I see some evil twins, –2x and 2x, so let’s add these two equations and get:

y = –2x + 4

y = 2x + 4

+_______________

2y = 0 + 8

Which we can solve: 2y = 8 y = 4. Plugging that into the first equation, we solve for x:

y = –2x + 4 4 = –2x + 4 0 = –2x x = 0. That would give us the

point (0, 4) as their intersection. Let’s test this in the other equation: y = 2x + 4

4 = 2(0) + 4 4 = 4? Yep! So (0, 4) is indeed the solution.

Time to graph! For both lines, we have the point (0, 4), which happens to be the

y-intercept for both of them, since it hits the y-axis.

For the first line, let’s plug in something easy like x = 1 and get: y = –2(1) + 4 y = 2,

so we get the point (1, 2). Next let’s plug in x = 2, and get: y = –2(2) + 4 y = 0, so we

get the point (2, 0); hey, that’s the x-intercept! Now we can graph this line.

For the second line, let’s plug in x = –2 and we get: y = 2(–2) + 4 y = 0, so we get the

point (–2, 0). Nice, that’s the x-intercept for this line. We’ll need one more point, how

about plugging in x = 1, and we get: y = 2(1) + 4 y = 6, so we get the point (1, 6).

Ready to graph the second line, and we’re done!

Answer:

Part a: consistent & independent

Part b: The solution is the point (0, 4)

Part c: See graph below

3. Let’s put that second line in slope-intercept form by dividing both sides by 5, and we

get: y = 5x. Notice that this and the first line both have a slope of 5. But they have

different y-intercepts (5 and 0, respectively), which means they are parallel and have no

solution! So they are inconsistent and there is no solution.

Now our only job is to graph the lines. For the first line, y = 5x + 5, we know the y-

intercept is (0, 5), and let’s find two more points: plugging in x = –1, we get y = –5 + 5 =

0, so the point (–1, 0), which is the x-intercept, and plugging in x = –2, we get y = –10 + 5

= –5 for the point (–2, –5). Ready to graph it!

For the next line, y = 5x, we know the y-intercept (and x-intercept, for that matter!) is

(0, 0). For more points, let’s plug in x = 1, and we get y = 5; the point (1, 5). And

plugging in x = –1, we get y = –5 for the point (–1, –5). Now we have enough points to

graph it. Done!

Answer:

Part a: inconsistent

Part b: no solution

Part c: See graph below

4. Let’s put the first line in slope-intercept form, by first multiplying both sides by 2, and

we get: x =y2− 3 2(x) = 2

y2− 3⎛

⎝⎜⎞⎠⎟ 2x = y – 6, and then adding 6 to both

sides, we get: 2x + 6 = y , in other words, y = 2x + 6. Hey wait a minute, this is the same

as the second equation! That means they are the same line, and so they are consistent &

dependent. It also means that the “solution” to this pair of equations is the entire line

itself: y = 2x + 6.

Now, although we can label it with both equations, we really only have one line to graph!

We can see from its equation that its y-intercept is (0, 6), and let’s get some other points

by plugging in x = –1 and we get y = 2(–1) + 6 y = 4 for the point (–1, 4), and

plugging in x = –2 we get y = (–2)2 + 6 y = 4 – 6 = 2, for the point (–2, 2). Done!

Answer:

Part a: consistent & dependent

Part b: The solution is the entire line: y = 2x + 6

Part c: See graph below

5. Putting the lines in slope-intercept form by solving for y, we’ll get: y = −12x + 5 for

the first line and y = 3x – 1 for the second line. They have different slopes, so they’ll be

consistent & independent, crossing at a single point, which will be our solution! To find

that solution, let’s go back to the original form of the first equation: x + 2y = 10 because

it’s always nice to avoid fractions, isn’t it? Looks like substitution should work pretty

well to find the solution. Using our newly rewritten second equation, y = 3x – 1, we’ll

plug (3x – 1) where we see y in the first equation:

x + 2y = 10 x + 2(3x – 1) = 10 x + 6x – 2 = 10 7x = 12 x =127

.

Okay, now let’s plug that into one of the equations, how about the second equation,

y = 3x – 1, and solve for y:

y = 3x – 1 y = 3127

⎛⎝⎜

⎞⎠⎟−1 y =

367

−77

y =297

. So that gives us the

point 127, 297

⎛⎝⎜

⎞⎠⎟ .

Hm, so much for avoiding fractions… but let’s plug this into the first original equation

and see if we can possibly get a true statement:

x + 2y = 10 127

+ 2 297

⎛⎝⎜

⎞⎠⎟= 10 ?

127+587

= 10 ? 707

= 10? 10 = 10. Yep!

So we’ve found our solution.

For part c, we’ll graph these lines by finding points – let’s not use the intersection point,

127, 297

⎛⎝⎜

⎞⎠⎟ , because it’s too hard to find those values!

For the first line, x + 2y = 10, let’s plug in x = 6, and we get: 6 + 2y = 10 y = 2, for the

point (6, 2). Let’s also plug in x = 2 and get 2 + 2y = 10 2y = 8 y = 4, for the point

(2, 4), and finally we’ll plug in x = –2 and get –2 + 2y = 10 2y = 12 y = 6, for the

point (–2, 6). Now we have enough points to graph that line.

For the second line, y = 3x – 1, we see the y-intercept is (0, –1), and let’s plug in x = 1 to

get y = 3(1) – 1 y = 2 for the point (1, 2), and plug in x = 2 to get y = 3(2) – 1 y = 5

for the point (2, 5). Ready to graph!

Taking a look at where these two lines intersect… hm, 127

should be a little smaller than

126

, in other words, a little smaller than 2, right? And 297

should be a little bigger than

287

, in other words, it’s a little bigger than 4. Yep, looking at where they intersect,

127, 297

⎛⎝⎜

⎞⎠⎟ seems about right! We’ll label their intersection point, and then we’re done!

Answer:

Part a: consistent & independent

Part b: 127, 297

⎛⎝⎜

⎞⎠⎟

Part c: See graph below

6. In the previous problem, we put this first equation in slope-intercept form, and we got:

y = −12x + 5 (just by subtracting x from both sides and dividing both sides by 5). The

second line, y = 3, has a slope of zero, so these two slopes are definitely different! That

means the two lines are consistent & independent; they cross at a single point. Let’s

find out what that point is!

Well, for the line y = 3, the y-value is ALWAYS 3, but for the first line, it will only have

one point that has a y-value of 3, and we just need to find out what that one point is.

Make sense? You can also just think of it as a regular substitution problem, where we

substitute 3 for y in the first equation. Let’s use the original form of the first equation:

x + 2y = 10 x + 2(3) = 10 x + 6 = 10 x = 4.

So that’s the intersection point (4, 3). After all, that point is on both lines isn’t it? And

they only cross at one point, so that has to be the one.

Okay, we’re ready to graph them. For the first line, we have the point (4, 3), and the

y-intercept is (0, 5). So let’s find one more point. Plugging in x = 2, we get:

y = −12x + 5 y = −

12(2) + 5 y = –1 + 5 y = 4, for the point (2, 4)

Okay, ready to graph that line.

And the line y = 3 is easy to graph, just pick points like (0, 3), (1, 3), (–4, 3), etc!

Answer:

Part a: consistent & independent

Part b: (4, 3)

Part c: See graph below, on next page