# System Of Equations

## Solving By Graphing

To solve by graphing you must first turn your equation into slope-intercept.

For this equation you must move the x variable to the opposite side.

Your new equations will be

7y=-2x+14

7y=-5x-7

After getting the y to be by it self, you then need to divide everything by the y variables coefficient. After dividing each side, your new equations will be

y=-2/7x+2

y=-5/7x-1

You then will graph these equations, like on the picture bellow this. ## Solving By Substitution

For this equation, you must "dress" it. Meaning you have to use one of the equation, and single out a variable, and make it's coefficient one. This one has x with a coefficient of one, so I will add negative 4y to each side.

x=10-4y

Now that you have x=10-4y, you have to substitute it into the other equation, 3x-5y=13.

3(10-4y)-5y=13

30-12y-5y=13

30-17y=13

Subtract 30 from each side

-17y=-17

Divide each side by -17

y=1

Now that you have y, you have to put it into both equations to get x, make sure x is the same number for both.

3x-5(1)=13

3x-5=13

Add 5 to each side

3x=18

Divide each side by 3

x=6

Now use this x and y for the other equation

1(6)+4(1)=10

6+4=10

10=10

The answer for this set of equations is x=6 and y=1. (6,1)

## Solving By Elimination

Just like the previous way to solve the equations, you must "dress" it. For this equations the x and y variables must be on the same side. The new equations for the question will be

-x+3y=5

-3x+8y=8

After doing this step you must make one of the variables equal zero, for example, x+-x=0x, or -5x+5x=0x.

For this set of equations I'll multiply the first equation by negative 3.

(-x+3y=5)-3

3x-9y=-15

Now add this to the other equation.

3x-9y=-15

+-3x+8y=8

You will get

-y=-7

This means y=7.

Now you put y=7 back into the original equations.

3(7)=x+5

21=x+5

Subtract 5 from each side

16=x

For the other equation

-3(16)+8(7)=8

-48+56=8

8=8

The answer for this set of equation is x=16 and y=7. (16,7)

## When should I use Substitution? When should I use Elimination? When should I use Graphing?

You should use Substitution whenever the question has the an equation that already has a variable isolated with a coefficient of 1.

You should use Elimination whenever you have a question that has the x and y variable on the same side of the equation for both equations.

You should use graphing whenever you have the y variable isolated with and the coefficient is one for both variables.

## Types of Solutions

There are there three types of solutions.

There is No Solution which happens when there is not possible answer. On a graph the would never intersect, they would always be parallel.

There is Infinite Solutions which is when the line always intersect. On a graph they would be a single line.

There is One Solution which is when there is a single answer, it intersects at a single point. On a graph it would be two lines intersecting only for one point of the line.

## Solving By Writing The System Of Equations

To solve this type of question, you need to create the equations. For this question the equation will be

24.50=10g+1q

22=8g+2q

In this equation, I will use elimination to solve it.

(24.50=10g+1q)-2

-49=-20g-2q

+22=8g+2q

-27=-12g

Divde each side by -12

2.25=g

Putting this into the original equations

24.50=10(2.25)+q

24.50=22.50+q

Subtract 22.50 from each side

2=q

Now try this with your other equations.

22=8(2.25)+2(2)

22=18+4

22=22

This means each quart is 2 dollars,and each gallon is 2.25 dollars. (2.25, 2) (g,q)

## Solving Word Problems

Like writing an equation, you create the equations.

x+4y=63

x+6y=87

Like the previous question, I will use elimination.

(x+4y=63)-1

-x-4y=-63

+x+6y=87

2y=24

Divide by 2 on each side

y=12

Now use this for the original equations.

x+4(12)=63

x+48=63

Subtract 48 from each side

x=15

Now use this for the other equation

1(15)+6(12)=87

15+72=87

87=87

This means there is a flat fee of 15 dollars, and a charge of 12 dollars per hour. (15,12)