System Of Equations

By: Eric D.

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)