Transcript Tutorial 6d - C on T ech Math : : An application

Solving Equations with
Variables on Both Sides
Tutorial 6d
Introduction
 In the previous lessons you learned how to solve one
and two step equations, like:

4x = 12 or 7x – 5 = 16
 In this lesson you will learn to solve equations with
variables on both sides, for example:

3x – 5 = 5 + 2x or 4(2x + 1) = 6 – 5x
 The steps in solving an equation that has variables on
both sides is similar to that of two-step equations,
except you must get the variable on just one side first.
Example #1
 Solve: 2x + 3x – 3 = 2x + 12
 Combine like terms.
 The goal is to get the variable on
just one side of the equation.
 That can be done by subtracting a
2x from both sides of the equation.
 Now it is a two-step equation. To
undo the subtraction, add a 3 to each
side of the equation.
 To undo the multiplication, divide
by 3 on each side of the equation.
5x – 3 = 2x + 12
-2x
-2x
3x – 3 = 12
+3 +3
13x = 15 5
3
3 1
1
x = 5
Always check your answers!
2x + 3x – 3 = 2x + 12; Does x =5?
2•5 + 3•5 – 5 = 2•5 + 12 is true!
Example #2
 Solve: 3x – 40 = 2(x – 5)
 Distribute the 2 on the right side.
 The goal is to get the variable on
just one side of the equation.
 That can be done by subtracting a
2x from both sides of the equation.
 Now it is a one-step equation. To
3x – 40 = 2x – 10
-2x
-2x
x – 40 = -10
+ 40 + 40
x = 30
undo the subtraction, add a 40 to
each side of the equation.
Always check your answers!
3x – 40 = 2(x – 5); Does x =30?
3•30 – 40 = 2(30 – 5) is true!
Problem Solving
 Mary & Jocelyn are sisters. They left school at 3:00 P.M. and
bicycled home along the same bike path. Mary rode at a speed
of 12 mi/h. Jocelyn rode at 9 mi/h. Mary got home 15 minutes
before Jocelyn. How long did it take Mary to get home?
Define your variable: Let t = Mary’s time in hours
Let t + .25 = Jocelyn’s time in hours.
Since Mary got home 15 minutes before Jocelyn, Jocelyn
must have been traveling for 15 minutes longer than Mary.
15 minutes equals 1/4 or .25 of an hour.
Problem Solving
 Mary & Jocelyn are sisters. They left school at 3:00 P.M. and
bicycled home along the same bike path. Mary rode at a speed
of 12 mi/h. Jocelyn rode at 9 mi/h. Mary got home 15 minutes
before Jocelyn. How long did it take Mary to get home?
Define your variable: Let t = Mary’s time in hours
Let t + .25 = Jocelyn’s time in hours.
We will need to use the formula:
Distance = Rate • Time
= r • t
You can see that they are both d
traveling the same distance.
Mary’s distance
Jocelyn’s distance
Problem Solving
 Mary & Jocelyn are sisters. They left school at 3:00 P.M. and
bicycled home along the same bike path. Mary rode at a speed
of 12 mi/h. Jocelyn rode at 9 mi/h. Mary got home 15 minutes
before Jocelyn. How long did it take Mary to get home?
Define your variable: Let t = Mary’s time in hours
Let t + .25 = Jocelyn’s time in hours.
Relate: Mary’s Distance
(rate•time)
Write:
Solve:
12t
equals
Jocelyn’s distance
(rate•time)
=
9(t + .25)
12t = 9 (t + .25)
Problem Solving
 Mary & Jocelyn are sisters. They left school at 3:00 P.M. and
bicycled home along the same bike path. Mary rode at a speed
of 12 mi/h. Jocelyn rode at 9 mi/h. Mary got home 15 minutes
before Jocelyn. How long did it take Mary to get home?
 Distribute the 9 on the right side.
 The goal is to get the variable on
just one side of the equation.
 That can be done by subtracting a
9t from both sides of the equation.
 Now it is a one-step equation. To
undo the multiplication, divide each
side of the equation by 3.
12t = 9(t + .25)
12t = 9t + 9(.25)
12t = 9t + 2.25
-9t -9t
13t = 2.25 .75
3
31
1
t = .75
Problem Solving
 Mary & Jocelyn are sisters. They left school at 3:00 P.M. and
bicycled home along the same bike path. Mary rode at a speed
of 12 mi/h. Jocelyn rode at 9 mi/h. Mary got home 15 minutes
before Jocelyn. How long did it take Mary to get home?
It took Mary .75 hour or
45 minutes to get home.
12t = 9(t + .25)
12t = 9t + 9(.25)
12t = 9t + 2.25
-9t -9t
13t = 2.25 .75
3
31
1
t = .75