PlosMathLearning

Equations

Linear Equations



Equations involving unknown variables are solved by balancing the left member and the right member.

Example 1: In the equation y + 6 = 17 (Let us call it example 1),

I know that the left member must equal 17 for it to balance with the right member.

This means y would be 11 because 11 + 6 = 17. That one is easy enough to do the math without a formal procedure.

The proper procedure in solving for y is to "isolate" y. This means we want y to stand by itself on one side of the equal sign.

Now, keep in mind that we want every thing to "balance" on either side of the equal sign.

This means whatever I do to the left hand side of the equation I must do to the same on the right hand side.

So, to isolate y and solve the equation, I must "move" the 6 to the other side of the equal sign.

To do this, I must subtracting 6 from both sides of the equation:

y + 6 = 17

y + 6 - 6 = 17 - 6

y + 0 = 17 - 6

y = 11 .

Example 2:

Solve 3x = 18

To get x to stand by itself we must divide both sides by 3.

3x3 = 183

Hence, x = 6

Example 3:

Solve 2x – 5 = 15

Step 1: Add 5 to both sides of the equation.

2x – 5 + 5 = 15 + 5

2x = 20

Step 2: Divide both sides of the equation by 2.

2x2 = 202

x = 10


Solving Fractional Equations


A fractional equation, also called radical equation, is any equation that has at least one fraction in it.

To solve a fractional expression you can follow these steps:

1. Find the least common denominator (LCD) of both sides of the equation.

2. Multiply both sides of the equation by the LCD. This will remove the fraction. You will now have a linear equation.

3. Solve the linear equation.

Example : Solve 2y - 25 = 4

The LCD is 5.

Multiplying both sides by the LCD we get

(5)2y - 25(5) = 4(5)

10y - 2 = 20

10y = 20 + 2

10y = 22

y = 2210

y = 225


Equations and Brackets


Brackets: The terms inside a bracket are always to be taken as a whole.

Example: Solve 6(y + 2) = 18

This will give you 6 times y plus 6 times 2 equals 18

6y + 12 = 18

6y = 18 - 12

6y = 6

y = 66

y = 1

Simultaneous Equations


Important facts to remember:

1. The term "simultaneous" means to happen at the same time.

2. Simultaneous equations consist of more than one unknown.

3. When we solve one pair of simultaneous equations the solutions obtained must satisfy both equations at the same time.

4. Having the expressions 3x and 2y, 3 is called the coefficient of x and 2 is called the coefficient of y.

Solving simultaneous equations

Two linear simultaneous equations can be solved by using the methods of elimination. The term "eliminate" means to get rid of.

When using this method the following steps must be are important:

1. Multiply one or both equations by suitable numbers to make the coefficients of one of the unknowns numerically equal.

2. Eliminate (get rid of) the unknown with equal coefficient by adding or subtracting the equations. The signs in both equations must be different.

3. Solve the equations in order to obtained the value of one unknown.

4. Substitute this value in one of the equations to find the value of the other unknown.

In all examples below, we first make x become the unknown with equal coefficient.


Example : Solve 3x + 4y = 20 and 2x + 3y = 14

Step 1: Multiply both equations by suitable numbers to make the coefficients of one of the unknowns numerically equal.

We will make the coefficient of x numerically even.

2(3x + 4y = 20)

-3(2x + 3y = 14)

6x + 8y = 40

-6x - 9y = -42

Step 2: Eliminate (get rid of) the unknown with equal coefficient by adding or subtracting the equations. The signs in both equations must be different.

When we add the two equations we are left with, -y = -2 Step 3: Solve the equations in order to obtain the value of the unknown.

-y = -2

y = 2

Step 4: Substitute this value in one of the equations to find the value of the other unknown.

By taking the equation 3x + 4y = 20,

When y = 2

3x + 4(2) = 20

3x + 8 = 20

3x = 20 - 8

3x = 12

x = 123

x = 4.



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