7.4 Using substitution (and algebra) – instead of graphing to
solve linear systems.
Look at this linear system:
3x + 5y = 6
&
x=-4
You CAN solve it by graphing, but there is a faster way.
Notice how you are TOLD what x is. So just … ‘plug it into’ the
other equation.
3x + 5y = 6
“plug” the -4 into the place of x above.
x=-4
rewrite it as:
3(-4) + 5y = 6 … and find out what y is.
3(-4) + 5y = 6
-12 + 5y = 6
5y = 6 + 12
5y = 18
y=
𝟏𝟖
𝟓
… So the solution is: (-4,
𝟏𝟖
)
𝟓
… then verify (normally)
So… if you can rearrange one equation to be x= or y= … then
just substitute and avoid graphing.
Example 1 page 418: solving by substitution
Solve this: 2x - 4y=7
&
4x+y=5
Answer: Find the equation with a variable having a coefficient of
1 and rearrange.
Remember: 4x+y=5 means
4x+1y=5
So rearrange to get y= …
y= 5 - 4x
Then substitute it INTO the other equation
… and solve.
2x-4y=7
2x – 4(5 – 4x) = 7
2x – 20 + 16x = 7
18x – 20 = 7
18x = 27
x = 27/18
x = 3/2
x = 3/2 … = 1.5
Now you know x, but don’t forget to calculate what y is!

4x+y=5
4(1.5) + y = 5
 get y = -1
Thus (1.5, -1) is your solution.
Verify by plugging (1.5, -1) into the OTHER equation … or by
graphing.
NOTE 1: If you don’t have anything with a coefficient of 1 …
… MAKE one of them into 1 by dividing.
e.g. 2x - 4y=7
&
4x+2y=5
rearrange
4x = 5 – 2y
÷4
÷4
÷4
… divide all by 4
1x = 1.25 – 0.5y
Then substitute and solve
𝟐𝒙 − 𝟒𝒚 = 𝟕
 I put as decimal for ease of calc.
𝟐(𝟏. 𝟐𝟓 – 𝟎. 𝟓𝒚) – 𝟒𝒚 = 𝟕
𝟐. 𝟓 – 𝟏𝒚 – 𝟒𝒚 = 𝟕
𝟐. 𝟓 – 𝟓𝒚 = 𝟕
– 𝟓𝒚 = 𝟕 – 𝟐. 𝟓
– 𝟓𝒚 = 𝟒. 𝟓
𝒚 = − 𝟒. 𝟓/𝟓
… so find x then state solution. Etc.
NOTE 2: OR INSTEAD … From same system of equations
… make one of them into the ‘other’ by multiplying.
e.g.
𝟐𝒙 − 𝟒𝒚 = 𝟕
&
𝟒𝒙 + 𝟐𝒚 = 𝟓
double entire equation
to make the 2x into 4x … to replace 4x here
𝟐(𝟐𝒙 − 𝟒𝒚 = 𝟕)
𝟐(𝟐𝒙) − 𝟐(𝟒𝒚) = 𝟐(𝟕)
𝟒𝒙 − 𝟖𝒚 = 𝟏𝟒
𝒎𝒖𝒔𝒕 𝒓𝒆𝒂𝒓𝒓𝒂𝒏𝒈𝒆 𝒕𝒐 𝒊𝒔𝒐𝒍𝒂𝒕𝒆 𝟒𝒙
𝟒𝒙 = 𝟏𝟒 + 𝟖𝒚
𝟒𝒙 = 𝟏𝟒 + 𝟖𝒚
fits into
now substitute and solve
𝟒𝒙 + 𝟐𝒚 = 𝟓
(𝟏𝟒 + 𝟖𝒚) + 𝟐𝒚 = 𝟓
𝟏𝟒 + 𝟏𝟎𝒚 = 𝟓
𝟏𝟎𝒚 = 𝟓 − 𝟏𝟒
𝟏𝟎𝒚 = −𝟗
−𝟗
𝒚=
𝟏𝟎
… so find x then state solution. Etc.
… choose the method easiest for you.
→ 𝟒𝒙 + 𝟐𝒚 = 𝟓
Example 2 page 420: Making your own equations
Nuri invested $2000 total last year. Some of it went at 8% annually
and the rest at 10%. After 1 year she had $190 in interest.
(a) Create a linear system out of this.
(b) How much money did she invest at each rate?
Answer:
Let e stand for the amount at 8% and let t stand for amount at
10%.
AMOUNT totalled $2000 so
… AMOUNT e + AMOUNT t = $2000
So…
Then:
e + t = $2000
8% times e … plus … 10% times t = $190
so…
0.08e + 0.10t = $190
Now solve by substituting:
e + t = $2000

0.08e + 0.10t = $190
rearrange
e = 2000 - t
substitute
0.08(2000 – t) + 0.10t = 190
160 – 0.08t + 0.10t = 190
160 + 0.02t = 190
collect like terms
subtract 160 from both sides
0.02t = 190 – 160
0.02t = 30
divide both sides by 0.02
t = 1500
Thus $1500 at 10% and the rest ($500) at 8%.
Example 3 page 422: What if you have FRACTIONS for
coefficients?
Solve the following by substitution:
𝟏
𝒙
𝟐
𝟐
+ 𝟑 𝒚 = −𝟏
&
𝟏
𝟓
𝒚 = 𝟒𝒙 − 𝟑
Answer: You can either KEEP the fractions then work it out by
substitution AS FRACTIONS …
Like so:
𝟏
𝒙
𝟐
𝟐
sub
+ 𝟑 𝒚 = −𝟏
𝟏
𝒙
𝟐
𝟐 𝟏
𝟓
+ 𝟑 (𝟒 𝒙 − 𝟑) = −𝟏
… or you can multiply by LCM in each to get rid of fractions.
(might be faster and easier in long run.)
This is called making an equivalent linear system
e.g.
𝟏
𝒙
𝟐
𝟐
+ 𝟑 𝒚 = −𝟏
 ALL times 6
𝟏
𝟐
𝟔 ( 𝒙 + 𝒚 = −𝟏)
𝟐
𝟑
𝟏
𝟐
𝟔 ( 𝒙) + 𝟔 ( 𝒚) = −𝟏(𝟔)
𝟐
𝟑
You end up with:
𝟑𝒙 + 𝟒𝒚 = −𝟔
Then ‘fix’ the other equation, too, if necessary:
𝟏
𝟓
Multiply everything by 12
(Doesn’t need to be same
# as you multiplied by before)
𝒚 = 𝟒𝒙 − 𝟑
𝟏
𝟓
𝟏𝟐𝒚 = 𝟏𝟐 ( 𝒙) − 𝟏𝟐 ( )
𝟒
𝟑
𝟏𝟐𝒚 = 𝟑𝒙 − 𝟐𝟎
Now you can substitute as normal.
𝟏𝟐𝒚 = 𝟑𝒙 − 𝟐𝟎
Notice that you have 3x in this equation
and other equation can be rewritten to
say what 3x equals.
3x = -6 – 4y
𝟏𝟐𝒚 = (−𝟔 − 𝟒𝒚) − 𝟐𝟎
𝟏𝟐𝒚 = −𝟔 − 𝟒𝒚 − 𝟐𝟎
𝟏𝟐𝒚 = −𝟐𝟔 − 𝟒𝒚
𝟏𝟐𝒚 + 𝟒𝒚 = −𝟐𝟔
𝟏𝟔𝒚 = −𝟐𝟔
𝒚=
−𝟐𝟔
𝟏𝟔
… lowest terms! →
𝒚=
−𝟏𝟑
𝟖
Now just find your x-value…by putting your y-value into one of
the equations. (You can use ANY of the equations … use the
easiest one)
𝟏𝟐𝒚 = 𝟑𝒙 − 𝟐𝟎
𝟏𝟐 (
−𝟏𝟑
) = 𝟑𝒙 − 𝟐𝟎
𝟖
−𝟑𝟗
= 𝟑𝒙 − 𝟐𝟎
𝟐
−𝟑𝟗
+ 𝟐𝟎 = 𝟑𝒙
𝟐
−𝟑𝟗 𝟒𝟎
+
= 𝟑𝒙
𝟐
𝟐
𝟏
= 𝟑𝒙
𝟐
𝟏 𝟏
÷ =𝒙
𝟐 𝟑
𝟏
=𝒙
𝟔
𝟏
−𝟏𝟑
𝑺𝒐 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝒕𝒉𝒆 𝒔𝒚𝒔𝒕𝒆𝒎 𝒊𝒔 𝒙 = 𝒂𝒏𝒅 𝒚 =
𝟔
𝟖
Check and verify.
Assignment:
CYU 1 – 3 pages 418 - 422
Then pages 423 – 426
[4ab, 5ab, 6, 8, 9, 10, 11, 12, 15, 16, 19]