3.2- Solving an Equation (Ax + B = C)

3.2- Solving an equation (ax + b = c),

Fractions

Curriculum Outcomes:

A6apply properties of numbers when operating upon expressions and equations

B1model (with concrete materials and pictorial representations) and express the relationships between arithmetic operations and operations on algebraic expressions and equations

C16interpret solutions to equations based on context

C27solve linear and simple radical and exponential equations and linear inequalities

Algebra Tiles

Algebra Tiles also known as Alge-Tiles are an asset to your classroom. These tiles can be used to show operations with integers, combining like terms, multiplication of polynomials, factoring of polynomials and solving equations. Many students are visual, hands-on learners. A visual or manipulative representation of algebraic expressions and operations can facilitate and enhance the understanding of many concepts that are presented in Math. The use of algebra tiles is present in various parts of this algebra supplement. The tiles are shown below.

The flip side of each tile is white and will be used to represent the negative values of the above tiles.

Solving Linear Equations

To solve a linear equation means to determine the value of the variable that makes the equation true. This can be done using algebra tiles (alge tiles). However, the process must be demonstrated by using equations for which you have enough tiles and for which fractional tiles are not needed. Fractional tiles do not exist.

Example 1: x + 5 = 3

Step 1: Model the equation using alge tiles.

Step 2: To isolate the variable x, apply the zero principle (take away 5 one tiles from each side of the equation).

Step 3: The remaining tiles show the value of x.

x = -2

Example 2: 4x + 3 = 2x - 9

Step 1: Model the equation using alge tiles.

Step 2: Subtract 2 x-tiles from each side of the equation so that all x-tiles are on one side.

Here is the result of step 2

Step 3: To isolate the variable (x), subtract three 1-tiles from each side.

Here is the result of step 3.

Step 4: Solve for x. We can see that 2x = -12. To find the value for x, we will split the tiles into 2 groups (2x).

So: x = -6

Example 3: 3(x – 1) = 2(x + 4)

Step 1: Model each side of the equation. 3 sets of x – 1 and 2 sets of x + 4.

Step 2: Subtract 2 x-tiles from each side of the equation so that all of the x-tiles are on one side.

The result of step 2 is:

Step 3: To isolate the variable (x), add three 1-tiles to each side of the equation.

The result of step 3 is:

x = 11

Exercises:

For each of the following equations, a solution is given. Determine if this given value is the solution to the equation.

EQUATION / SOLUTION
A / 2x – 5 = 11 / 6
B / 5(a + 3) = 0 / 0
C / 3e – 2 = e + 4 / 3
D / 3y – 5 = 2y / 5
E / 3(m – 2) = 2 – m / -2
F / /

Solve each of the following and check.
3(a + 4) – 2(a + 3) = 1
4(b + 7) – 2(b – 5) = 3(b – 2)
0.5(d + 0.7) – 0.2(d – 0.3) = 0.4
2(e + 4) + 3(e + 2) = -2e

Solve for x.
mx – n = a
3x(m – n) – n = a – m
2m(x – n) + a = n

Answers:

A. 2x – 5 = 11check: x = 6

2(6) – 5 = 11

12 – 5 = 11

7 ≠ 11

 6 is not the solution

B. 5(a + 3) = 0check: x = 0

5(0 + 3) = 0

5(3) = 0

15 ≠ 0

 0 is not the solution

C. 3e – 2 = e + 4check: x = 3

3(3) – 2 = (3) + 4

9 – 2 = (3) + 4

7 = 7

 3 is the solution

D. 3y – 5 = 2ycheck: x = 5

3(5) – 5 = 2(5)

15 – 5 = 10

10 = 10

 5 is the solution

E. 3(m – 2) = 2 – mcheck: x = -2

3((-2) – 2) = 2 – (-2)

3(-4) = 2 + 2

-12 ≠ 4

 -2 is not the solution

F. check: x =

 is not the solution

a. 3(a + 4) – 2(a + 3) = 1

3a + 12 – 2a – 6 = 1

a + 6 = 1

a + 6 – 6 = 1– 6

a = -5

The solution is -5.

b.4(b + 7) – 2(b – 5) = 3(b – 2)

4b + 28 – 2b + 10 = 3b – 6

2b + 38 = 3b – 6

2b – 3b + 38 = 3b – 3b – 6

-b + 38 – 38 = -6 – 38

-b = -44

b = 44The solution is 44.

1(3c – 3) + 2(-c – 3) = 1(1)

3c – 3 – 2c – 6 = 1

c – 9 = 1

c – 9 + 9 = 1 + 9

c = 10

The solution is 10

0.5(d + 0.7) – 0.2(d – 0.3) = 0.4

0.5d + 0.35 – 0.2d + 0.06 = 0.4

0.3d + 0.41 = 0.4

0.3d + 0.41 – 0.41 = 0.4 – 0.41

0.3d = -0.01

d = -0.03

The solution is -0.03.

2(e + 4) + 3(e + 2) = -2e

2e + 8 + 3e + 6 = -2e

5e + 14 = -2e

5e + 2e + 14 = -2e + 2e

7e + 14 – 14 = 0 – 14

7e = -14

e = -2 The solution is -2.

a. mx – n = a

mx – n + n = a + n

mx = a + n

3x(m – n) – n = a – m

3x(m - n) – n + n = a – m + n

3x(m – n) = a – m + n

2m(x – n) + a = n

2m(x – n) + a – a = n – a

2m(x – n) = n– a