Algebra ( Period 1)

SOLVING EQUATIONS WITH VARIABLE ON BOTH SIDES 2-4 

COLLECTING TERMS FIRST:
Sometimes, you will have to COLLECT LIKE TERMS ON THE SAME SIDE OF THE EQUATION before balancing:

8y + 12 – (-2y) = -6

10y + 12 = -6

10y = -18

y = -18/10 =  -9/5

TWO STEPS WITH DISTRIBUTIVE PROPERTY
Usually, you want to do DISTRIBUTE FIRST!
UNLESS THE FACTOR OUTSIDE THE ( ) CAN BE DIVIDED
OUT OF BOTH SIDES PERFECTLY!!!!

EXAMPLE:
5y - 2(2y + 8) = 16
5y - 4y - 16 = 16 [distribute]
y - 16 = 16 [collect like terms]
y = 32 [solve by adding 16 to both sides]

EXAMPLE WHEN YOU DON'T NEED TO DISTRIBUTE FIRST:
-3(4 + 3x) = -9
4 + 3x = 3 [Don't distribute! Divide by -3. The -3 goes into both sides perfectly!)
3x = -1 [Subtract 4 from both sides]
x = -1/3 [Divide both sides by 3]

REVIEWED: IDENTITY OR NO POSSIBLE SOLUTION EQUATIONS:

An identity equation is where ANY NUMBER can be substituted for the VARIABLE, the equation will be TRUE. What will happen is that while you’re balancing the equations, you will ultimately end up with the SAME EXACT EXPRESSION ON EACH SIDE of the equation.

 

You can keep going, but as soon as you have the same thing on both sides, you know you have an IDENTITY

A no possible solution equation is one where no matter what number you substitute into the equation, the equation will be FALSE. What will happen is that while you’re balancing the equations, you will ultimately end up with one number will equal a DIFFERENT number (which can never be true).

 

NOTICE SOMETHING ELSE ABOUT SOLVING EQUATIONS IN GENERAL:

WHENEVER YOU HAVE THE SAME EXACT TERM WITH THE SAME SIGN ON DIFFERENT SIDES OF THE EQUATION, YOU CAN SIMPLY CROSS THEM OUT BECAUSE  WHEN YOU USE THE ADDITIVE INVERSE PROPERTY ON BOTH SIDES TO BALANCE, BOTH TERMS WILL DROP OUT!

VARIABLES ON BOTH SIDES:

Simplify each side of the equation first.
Then use the ADDITIVE INVERSE PROPERTY to move variables to the other side of the equation so that all variables are on the same side.
Usually, we try to move the smaller coefficient to the larger because sometimes that avoids negative coefficients,

BUT that is not always the case, and you may move to whatever side you choose!
EXAMPLE:
3y - 10 - y = -10y + 12
2y - 10 = -10y + 12
+10y +10y
12y - 10 = 12
+ 10 +10
12y = 22
12     12
y = 11/6

FINDING THE VALUE OF AN UNKNOWN SO THAT 2 PERIMETERS OR AREAS ARE THE SAME:

This type of problem is a perfect example of using the Distributive Property with variable on both sides of an equation.

Example 4 on p. 99

You have 2 rectangles whose areas are the same. One rectangle has sides of x and 10 cm. and the other rectangle has sides of x + 3 and 6 cm.

Set up the following equation and solve:

10x = 6(x + 3)

10x = 6x + 18

4x = 18

x = 18/4 = 9/2 cm

CHECK THAT THE RECTANGLES DO HAVE THE SAME AREA IF x = 9/2 OR 4.5 cm

First rectangle: Sides are 4.5 and 10 so A = (4.5)(10) = 45 cm2

Second rectangle: Sides are 4.5 + 3 or 7.5 cm and 6 cm. so A = (7.5)(6) = 45 cm2

You have 2 rectangles whose perimeters are the same. One rectangle has sides of x and 6 cm. and the other rectangle has sides of 2x + 2 and x cm.

Set up the following equation and solve:

2x + 2(6) = 2(2x + 2) + 2x

2x + 12 = 4x + 4 + 2x

2x + 12 = 6x + 4

12 = 4x + 4

8 = 4x

x = 2

CHECK THAT THE RECTANGLES DO HAVE THE SAME PERIMETER IF x = 2 cm

First rectangle: Sides are 2 and 6 so P = 2(2) + 2(6) = 16 cm

Second rectangle: Sides are 2 and 2(2) + 2 = 6 so P = 2(2) + 2(6) = 16