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
get: 2y + 8x = 2 
1
1
( 2y + 8x ) = (2)
2
2

1
and
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 =
1
4
Okay, now we’ll plug this into one of the equations – how about 2x – 4y = 3, and we get:
2x – 4y = 3

⎛ 1⎞
2x − 4 ⎜ ⎟ = 3
⎝ 4⎠
 2x – 1 = 3
And let’s test these values, x = 2 and y =
found the right values: y =
Answer: x = 2, y =
1
4
x

8

2x = 4 
x=2
1
, in the other original equation to see if we
4
1 2
= ?
4 8

1 1
Yep! Done!
=
4 4
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
−48
−10
y=

 y=
24
5
Okay, now we can plug this into either equation, how about 3x + 5y = 12, and get:
3x + 5y = 12

⎛ 24 ⎞
3x + 5 ⎜ ⎟ = 12
⎝ 5⎠

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
⎛ 24 ⎞
3(–4) − 5 ⎜ ⎟ = −36 ?
⎝ 5⎠
 –12 – 24 = –36? 
–36 = –36? Yep!
Answer: x = –4, y =
24
5
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)
and: 8x + 15y = 23 

2(8x + 15y) = 2(23) 
21x + 30y = 51
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

and 3x + 2y = –6

2(4x + 3y) = 2(–7)
 3(3x + 2y) = 3(–6)

8x + 6y = –14
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 yintercept 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 =
y
−3
2

⎛y
⎞
2(x) = 2 ⎜ − 3⎟
⎝2
⎠
 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
1
5. Putting the lines in slope-intercept form by solving for y, we’ll get: y = − x + 5 for
2
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 =
12
.
7
Okay, now let’s plug that into one of the equations, how about the second equation,
y = 3x – 1, and solve for y:
⎛ 12 ⎞
y = 3x – 1  y = 3 ⎜ ⎟ − 1
⎝ 7⎠

y=
36 7
−
7 7
 y=
29
. So that gives us the
7
⎛ 12 29 ⎞
point ⎜ , ⎟ .
⎝ 7 7⎠
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 
12
12 58
70
⎛ 29 ⎞
+ 2 ⎜ ⎟ = 10 ? 
= 10?  10 = 10. Yep!
+
= 10 ? 
⎝ 7⎠
7
7
7
7
So we’ve found our solution.
For part c, we’ll graph these lines by finding points – let’s not use the intersection point,
⎛ 12 29 ⎞
⎜⎝ , ⎟⎠ , because it’s too hard to find those values!
7 7
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,
12
should be a little smaller than
7
12
29
, in other words, a little smaller than 2, right? And
should be a little bigger than
6
7
28
, in other words, it’s a little bigger than 4. Yep, looking at where they intersect,
7
⎛ 12 29 ⎞
⎜⎝ , ⎟⎠ seems about right! We’ll label their intersection point, and then we’re done!
7 7
Answer:
Part a: consistent & independent
⎛ 12 29 ⎞
Part b: ⎜ , ⎟
⎝ 7 7⎠
Part c: See graph below
6. In the previous problem, we put this first equation in slope-intercept form, and we got:
1
y = − x + 5 (just by subtracting x from both sides and dividing both sides by 5). The
2
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:
1
y=− x+5
2

1
y = − (2) + 5
2

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