Basic math? Simply put, you use basic mathematics almost every day of your life. You use it at home, on the job,or when you go to school.

At home, you may for instance have a budget to help manage your income and probably put some money aside. If your monthly income is 1000 dollars, 50% may go to your rent or mortgage;

20% may go to food, clothing,gas, and other utilities;20% may be used for personal items, gifts, and entertainment; finally, the remaining will go to your saving account. What percentage is left for saving by the way?

The situation described above is a perfect example of basic skills in mathematics used at home. Believe it or not, you may need many skills in math to keep a budget up to date.

At home, you may for instance have a budget to help manage your income and probably put some money aside. If your monthly income is 1000 dollars, 50% may go to your rent or mortgage;

20% may go to food, clothing,gas, and other utilities;20% may be used for personal items, gifts, and entertainment; finally, the remaining will go to your saving account. What percentage is left for saving by the way?

The situation described above is a perfect example of basic skills in mathematics used at home. Believe it or not, you may need many skills in math to keep a budget up to date.

**Example #1**

The sum of two numbers is twenty and their difference is ten. What are the two numbers?

Here is how to set up the system:

Let x be the first number

Let y be the second number

Then,

x + y = 20

x − y = 10

You can also write:

x + y = 20

y − x = 10

You will get the same answers except that the values for x and y will be swapped

**Example #2**

You have 24 coins in your pockets that are worth 4.50 dollars. How many coins are quarters? How many coins are dimes?

Here is how to set it up:

Let q be the number of quarters

Let d be the number of dimes

Then,

q + d = 24

25 × q + 10 × d = 450

The second equation is tricky. How did we get it?

Since 1 quarter equal to 25 cents, q quarter equal to 25 × q

If you had 6 quarters and you wanted to know how many cents are there for the 6 quarters, would you not do 6 × 25?

Just say to yourself that now instead of 6 quarters you have q quarters. Does that make sense?

In a similar way, since 1 dime equal to 10 cents, d dimes equal 10 × d

What about the 450? 4.50 dollars times 100 = 450 cents

Finally, since 25 × q represents how many cents you have for quarters and 10 × d represents how many cents you have for dimes, adding them should equal to the total of 450 cents

**Example #3**

A cell phone plan offers 300 free minutes for a flat fee of 20 dollars. If your usage exceed 300 minutes, you pay 50 cents for each minute.

A second cell phone plan offers 500 free minutes for a flat fee of 30 dollars. If your usage exceed 400 minutes, you pay 30 cents for each minute

Model the cost of both plan with a system.

Here is how to set up the system:

Let x be the number of minutes you talk

Let y be the cost

y = 20 + 0.50 x

y = 30 + 0.30 x

Minus 0.50x from both sides in the first equation . Minus 0.30 x from both sides in the second equation

We get:

y − 0.50 x = 20

y − 0.30 x = 30

Now go to the lessons below to learn how to solve a system of linear equations. I have also included a system of linear equations solver

Substitution method

System of linear equations can also be solved using the substitution method.We will show with examples.

Before you learn this lesson, make sure you understand how to solve linear equations

x + y = 20

x − y = 10

You have two equations. Pick either the first or the second equation and solve for either x or y

Since I am the one solving it, I have decided to choose the equation at the bottom (x − y = 10) and I will solve for x

x − y = 10

Add y to both sides

x − y + y = 10 + y

x = 10 + y

Since you used the equation at the bottom to solve for x, you will substitute x into the equation on top (x + y = 20)

Using x + y = 20, erase x and write 10 + y since x = 10 + y

We get 10 + y + y = 20

10 + 2y = 20

Minus 10 from both sides

10 − 10 + 2y = 20 − 10

2y = 10

Divide both sides by 2

y = 5

Now you have y, you can replace its value into either equation to get x

Replacing y into x + y = 20 gives

x + 5 = 20

Minus 5 from both sides

x + 5 − 5 = 20 − 5

x = 15

The solution to the system is x = 15 and y = 5

Indeed 15 + 5 = 20 and 15 − 5 = 10

3x + y = 10

-4x − 2y = 2

You have two equations. Pick either the first or the second equation and solve for either x or y

I have decided to choose the equation on top (3x + y = 10) and I will solve for y

3x + y = 10

Subtract 3x from both sides

3x − 3x + y = 10 − 3x

y = 10 − 3x

Since you used the equation on top to solve for y, you will substitute y into the equation at the bottom (-4x − 2y = 2)

Using -4x − 2y = 2, erase y and write 10 − 3x keeping in mind that there is a multiplication between 2 and y

We get -4x − 2 ×(10 − 3x ) = 2

-4x − 2 ×(10 − 3x ) = 2

-4x − 20 + 6x = 2 (After multiplying -2 by 10 and -2 by -3x)

2x − 20 = 2

Add 20 to both sides

2x − 20 + 20 = 2 + 20

2x = 22

Divide both sides by 2

x = 11

Now you have x, you can replace its value into either equation to get y

Replacing x into 3x + y = 10 gives

3 × 11 + y = 10

33 + y = 10

Minus 33 from both sides

33 − 33 + y = 10 − 33

y = -23

The solution to the system is x = 11 and y = -23

Indeed, 3 × 11 + -23 = 33 + -23 = 10 and -4 × 11 − 2 × -23 = -44 + 46 = 2

Y

System of linear equations can also be solved using the substitution method.We will show with examples.

Before you learn this lesson, make sure you understand how to solve linear equations

**Example #1:**Solve the following system using the substitution methodx + y = 20

x − y = 10

**Step 1**You have two equations. Pick either the first or the second equation and solve for either x or y

Since I am the one solving it, I have decided to choose the equation at the bottom (x − y = 10) and I will solve for x

x − y = 10

Add y to both sides

x − y + y = 10 + y

x = 10 + y

**Step 2**Since you used the equation at the bottom to solve for x, you will substitute x into the equation on top (x + y = 20)

Using x + y = 20, erase x and write 10 + y since x = 10 + y

We get 10 + y + y = 20

10 + 2y = 20

Minus 10 from both sides

10 − 10 + 2y = 20 − 10

2y = 10

Divide both sides by 2

y = 5

Now you have y, you can replace its value into either equation to get x

Replacing y into x + y = 20 gives

x + 5 = 20

Minus 5 from both sides

x + 5 − 5 = 20 − 5

x = 15

The solution to the system is x = 15 and y = 5

Indeed 15 + 5 = 20 and 15 − 5 = 10

**Example #2:**Solve the following system using the substitution method3x + y = 10

-4x − 2y = 2

**Step 1**You have two equations. Pick either the first or the second equation and solve for either x or y

I have decided to choose the equation on top (3x + y = 10) and I will solve for y

3x + y = 10

Subtract 3x from both sides

3x − 3x + y = 10 − 3x

y = 10 − 3x

**Step 2**Since you used the equation on top to solve for y, you will substitute y into the equation at the bottom (-4x − 2y = 2)

Using -4x − 2y = 2, erase y and write 10 − 3x keeping in mind that there is a multiplication between 2 and y

We get -4x − 2 ×(10 − 3x ) = 2

-4x − 2 ×(10 − 3x ) = 2

-4x − 20 + 6x = 2 (After multiplying -2 by 10 and -2 by -3x)

2x − 20 = 2

Add 20 to both sides

2x − 20 + 20 = 2 + 20

2x = 22

Divide both sides by 2

x = 11

Now you have x, you can replace its value into either equation to get y

Replacing x into 3x + y = 10 gives

3 × 11 + y = 10

33 + y = 10

Minus 33 from both sides

33 − 33 + y = 10 − 33

y = -23

The solution to the system is x = 11 and y = -23

Indeed, 3 × 11 + -23 = 33 + -23 = 10 and -4 × 11 − 2 × -23 = -44 + 46 = 2

Y

**ou should have noticed that the reason we call this method the substitution method is because after you have solve for a variable in one equation, you substitute the value of that variable into the other equation**

Elimination method

System of linear equations can also be solved using the elimination method.We will show with examples.

Before you learn this lesson, make sure you understand how to solve linear equations

Here are the steps to follow:

Try to eliminate a variable as you add the left sides and the right sides of the two equations

Set the sum resulting from adding the left sides equal to the sum resulting from adding the right sides

Solve for the variable that was not cancelled or eliminated

Use the answer found in step 3 to solve for the other variable by substituting this value in one of the two equations

x + y = 20

x − y = 10

Examine the two equations carefully.Then, you will try to eliminate or cancel a variable by adding the left sides (x + y and x − y).

However, since you are adding the left sides, you have to the right sides (20 and 10) of the two equations also

For the equations above, it turns out that it is easy to eliminate y while adding the left sides since x + y + x − y = x + x + y − y = x + x + 0 = 2x

The sum for the left sides is 2x and the sum for the right sides is 20 + 10 = 30

Setting them equal, we get 2x = 30

2x = 30 Solve for x by dividing both sides of this equation by 2

(2/2)x = 30/2

x = 15

You can substitute 15 for x in either x + y = 20 or x − y = 10 to get y

Choosing the first one, we get 15 + y = 20

Minus 15 from both sides to get y = 5

Now check yourself that the answer is still the same if you had chosen to substitute 15 for x in x − y = 10

3x + y = 10

-4x − 2y = 2

First, notice that nothing can be eliminated when adding the left sides since

3x + y + -4x − 2y = -1x + 3y

However, in 3x + y = 10, if I can turn y into 2y, I could elimnate y by adding 2y to -2y in -4x − 2y = 2

Therefore, turn y into 2y by multiplying the whole equation by 2.

2 × ( 3x + y = 10) gives the new equation 6x + 2y = 20

Adding the left side of this equation to the left side of -4x − 2y = 2 gives what you see below:

6x + 2y + -4x − 2y = 6x + -4x + 2y − 2y = 2x + 0 = 2x

Adding the right gives us 20 + 2 = 22

The sum for the left sides is 2x and the sum for the right sides is 20 + 2 = 22

Setting them equal, we get 2x = 22

2x = 22

Solve for x by dividing both sides of this equation by 2

(2/2)x = 22/2

x = 11

You can substitute 11 for x in either 3x + y = 10 or -4x − 2y = 2 to get y

Choosing the first one, we get 3 × 11 + y = 10

33 + y = 10

Minus 33 from both sides to get y = -23

System of linear equations can also be solved using the elimination method.We will show with examples.

Before you learn this lesson, make sure you understand how to solve linear equations

Here are the steps to follow:

**Step 1**Try to eliminate a variable as you add the left sides and the right sides of the two equations

**Step 2**Set the sum resulting from adding the left sides equal to the sum resulting from adding the right sides

**Step 3**Solve for the variable that was not cancelled or eliminated

**Step 4**Use the answer found in step 3 to solve for the other variable by substituting this value in one of the two equations

**Example #1:**Solve the following system using the elimination methodx + y = 20

x − y = 10

**Step 1**Examine the two equations carefully.Then, you will try to eliminate or cancel a variable by adding the left sides (x + y and x − y).

However, since you are adding the left sides, you have to the right sides (20 and 10) of the two equations also

For the equations above, it turns out that it is easy to eliminate y while adding the left sides since x + y + x − y = x + x + y − y = x + x + 0 = 2x

**Step 2**The sum for the left sides is 2x and the sum for the right sides is 20 + 10 = 30

Setting them equal, we get 2x = 30

**Step 3**2x = 30 Solve for x by dividing both sides of this equation by 2

(2/2)x = 30/2

x = 15

**Step 4**You can substitute 15 for x in either x + y = 20 or x − y = 10 to get y

Choosing the first one, we get 15 + y = 20

Minus 15 from both sides to get y = 5

Now check yourself that the answer is still the same if you had chosen to substitute 15 for x in x − y = 10

**Example #2:**Solve the following system using the elimination method3x + y = 10

-4x − 2y = 2

**Step 1**First, notice that nothing can be eliminated when adding the left sides since

3x + y + -4x − 2y = -1x + 3y

However, in 3x + y = 10, if I can turn y into 2y, I could elimnate y by adding 2y to -2y in -4x − 2y = 2

Therefore, turn y into 2y by multiplying the whole equation by 2.

2 × ( 3x + y = 10) gives the new equation 6x + 2y = 20

Adding the left side of this equation to the left side of -4x − 2y = 2 gives what you see below:

6x + 2y + -4x − 2y = 6x + -4x + 2y − 2y = 2x + 0 = 2x

Adding the right gives us 20 + 2 = 22

**Step 2**The sum for the left sides is 2x and the sum for the right sides is 20 + 2 = 22

Setting them equal, we get 2x = 22

**Step 3**2x = 22

Solve for x by dividing both sides of this equation by 2

(2/2)x = 22/2

x = 11

**Step 4**You can substitute 11 for x in either 3x + y = 10 or -4x − 2y = 2 to get y

Choosing the first one, we get 3 × 11 + y = 10

33 + y = 10

Minus 33 from both sides to get y = -23

**You should have noticed that the reason we call this method the elimination method is because the first thing you do is eliminate a variable**System of linear equations solver

This system of linear equations solver will help you solve any system of the form:

.

ax + by = c

dx + ey = f

Enter a,b, and c into the three boxes on top starting with a.

Enter d,e, and f into the three boxes at the bottom starting with d.

a b c

x y =

d e f

x y =

Note: Your system should be written in the form above. If not, rewrite the system:

Study this example carefully:

Write 5x − 4 = 2y in the form above

Minus 2y from both sides to get 5x − 2y − 4 = 2y − 2y

We get 5x − 2y − 4 = 0

Add 4 to both sides

5x − 2y − 4 + 4 = 0 + 4

5x − 2y = 4

Here a, b, and c are 5, -2, and 4 respectively

This system of linear equations solver will help you solve any system of the form:

.

ax + by = c

dx + ey = f

Enter a,b, and c into the three boxes on top starting with a.

Enter d,e, and f into the three boxes at the bottom starting with d.

**Hit calculate**a b c

x y =

d e f

x y =

Note: Your system should be written in the form above. If not, rewrite the system:

Study this example carefully:

Write 5x − 4 = 2y in the form above

Minus 2y from both sides to get 5x − 2y − 4 = 2y − 2y

We get 5x − 2y − 4 = 0

Add 4 to both sides

5x − 2y − 4 + 4 = 0 + 4

5x − 2y = 4

Here a, b, and c are 5, -2, and 4 respectively