# Applying Properties of Equality

In this lesson we will learn how to apply The Properties of Equality to Systems of Equations so that we can manipulate and solve them.

Review,

## Properties of Equality

Additive Property - if a = b is true, a + c = b + c is also true

Multiplicative Property - if a = b is true, ac = bc is also true

What we will learn in this lesson is how we can apply this method simultaneously to multiple equations in a Systems of Equations. Generally, we can think of the format as this:

If, a, b, c, d, e, f are all numbers, and x and y are our variables all in a true System of equations that looks like:

ax + by = c

dx + ey = f

We can multiply one equation or both by any number, say g, and it still remains true, for example

ax + by = c

(dx + ey = f ) g

The system remains true,

We can multiply both equations in the system by two different numbers, say g and h,

(ax + by = c) h

(dx + ey = f) g

The system remains true.

We can multiply one equation or both by any number, say g, and it still remains true, for example

ax + by = c

(dx + ey = f) g - [remembering we must multiply both sides of the equation by g]

[(dx + ey) g = (f ) g]

The system remains true,

We can multiply both equations in the system by two different numbers, say g and h,

(ax + by = c) h

(dx + ey = f) g

The system remains true.

The same is true for the Additive Property,

We can add any number to one equation or both, say g, and it still remain true, for example

ax + by = c

(dx + ey) + g = (f) + g

The system still remains true,

We can add any numbers to both equations in the system by two different numbers, say g and h,

(ax + by) + h = (c) + h

(dx + ey) + g =(f) + g

The system still remains true.

We can also apply the additive property to add or subtract the equations in the system from each other, as long as we equate both sides,

ax + by = c

dx + ey = f

For example, if we take the first equation and subtract the second equation, the system will still remain true, as we are staying consistent with the “equals sign”

ax + by = c

- (dx + ey = f)

Becomes

ax - (dx) + by - (ey) = c - (f)

The equation will still remain true, as we are following the properties of equality.

Let’s get into some examples.

## Question 1

This is is a multi-part question: We have the point (5, 1) as a solution to the system of x + 2y = 7 and 2y - 2x = -8

Firstly, Add both equations together, then see if (5, 1) still remains a solution.

x + 2y = 7

+(2y - 2x = -8)

x + (-2x) + 2y + (2y) = 7 + (-8)

Equating like terms gives us,

-x + 4y = -1

Now we sub in (5, 1),

-(5) + 4(1) = -1

-5 + 4 = -1

-1 = -1

As we can see, after adding these equations together, the solution still remains true, as we followed the correct properties of equality.

Next, multiply the first equation (x + 2y = 7) by 2, then check if (5, 1) is still a solution.

Multiply x + 2y = 7 by 2, remembering to multiply both sides,

2(x + 2y) = 2(7)

2x + 4y = 14,

Now sub in (5, 1)

2(5) + 4(1) = 14

10 + 4 = 14

14 = 14

(5, 1) is still a solution.

Finally, with the equation created after multiplying by 2, add the second equation and see if you can solve for one of the variables and then the other.

We multiplied (x + 2y = 7) by 2 to give us 2x + 4y = 14

Now add the second equation, 2y - 2x = -8

2x + 4y = 14

+(2y - 2x = -8)

2x + (-2x) + 4y + (2y) = 14 + (-8)

2x - 2x + 4y + 2y = 6

0 + 6y = 6,

Now we can solve for y!

Divide both sides by 6,

6y / 6 = 6 / 6

y = 1

Now for the other variable:

Put y = 1 into either of the equations, try x + 2y = 7,

x + 2(1) =7,

Subtract both sides by 2,

x + 2 - 2 = 7 - 2

x = 5

There we have the solution x = 5 and y = 1 (5, 1). Note this is the solution we were given!

This is what is possible when you correctly apply the Properties of Equality to Systems of Equations, we can rearrange and find the solution!

## Question 2

This question will be a walkthrough question for solving the system of equations below.

x + 3y = 2

3x - 2y = 6

First step to solving this, will be multiplying each equation by a number to make the like terms cancel out. There is a 3y on the top and -2y on the bottom, we can manipulate these along with their equations to cancel them out, to leave us with one variable.

Multiply the top equation by 2 and the bottom by 3.

Multiply out,

(x + 3y = 2 )2

(3x - 2y = 6 )3

2x + 6y = 4

9x -6y = 18

Note, 6y and -6y look like they’ll cancel out!

Next, add the two equations together.

2x + 6y = 4

+(9x -6y = 18)

2x + 11x + 6y -6y = 18 + 4

11x + 0y = 22

11x = 22

Divide both sides by 11,

11x / 11 = 22 / 11

x = 2

Finally, use the x-value from b) to find the value of y and hence a solution to the system.

Last step to solving this system is to sub in x = 2 into one of our equations (generally, pick the equation that looks like the one you'd be most comfortable solving)

Let’s sub x = 2 into x + 3y = 2,

(2) + 3y = 2

Subtract both sides by 2,

2 - 2 + 3y = 2 - 2

3y = 0

Divide both sides by 3,

3y / 3 = 0 / 3

y = 0,

So, the point (2, 0) is a solution to our system.

## Question 3

Find a solution to the system of equations below:

2x + y = 1

3x + 6y = -3

First, start by looking at the coefficients of our y-terms and x-terms in both equations, and decide which one looks simpler to manipulate and eventually cancel out,

We have 2x in the first and 3x in the second, if we wanted these terms to cancel out when we added the equations together, we would have to multiply the first by either 3 or -3 and the second equation by either -2 or 2.

Looking at the y-terms we have y in the first and 6y in the second. For these to cancel out when we added them together, we would just have to multiply the first equation by -6, so let’s choose this option,

Multiply the first equation by -6 and then add the equations together,

(2x + y = 1) -6

3x + 6y = -3

(-6)2x + (-6) y = (-6)1

3x + 6y = -3

-12x - 6y = -6

3x + 6y = -3

-12x - 6y = -6

+(3x + 6y = -3)

-12x + 3x -6y + 6y = -6 - 3

Notice, the -6y and 6y cancelling out as planned,

-9x = -9

Divide both sides by -9,

-9x / -9 = -9 / -9

x = 1

There we have the first part of a solution to the set, sub x = 1 into one of the equations and find the y-value, do 2x + y = 1

2(1) + y = 1,

Subtract both sides by 2

2 + y - 2 = 1 - 2

y = -1,

So, we have x = 1 and y = -1 (1, -1) as a solution to our system of equations. Become familiar with this technique and learn how to identify how to manipulate equations to be able to solve them.

## Question 4

Find the solution to:

2x + y = 7

-3x - 2y = -7

Notice, we have a y in the top equation and a -2y in the bottom, we can multiply the top by 2 to give us -2y and 2y which can cancel out our y terms,

2(2x + y = 7)

-3x - 2y = -7

2(2x) + 2(y) = 2(7)

-3x - 2y = -7

4x + 2y = 14

-3x - 2y = -7

4x + 2y = 14

+(-3x - 2y = -7)

4x + (-3x) + 2y + (-2y) = 14 + (-7)

x = 7,

Now sub that x value back into one our equations, try 2x + y = 7

2(7) + y = 7

14 + y = 7,

Subtract 14 from both sides,

14 + y - 14 = 7 - 14

y = -7,

So, our solution is, (7, -7) or x = 7 y = -7