There are a couple of ways to solve a system of linear equations: graphically and algebraically. In this lesson, we will review the two ways to solve a system of linear equations algebraically: substitution and elimination. 

Substitution can be used by solving one of the equations for either  or , and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the  form, and then set both equations equal to each other. 

Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable. 

For this problem, substitution makes the most sense because the first equation is already solved for a variable. We can substitute the expression that is equal to , into the  of our second equation:

Next, we need to distribute and combine like terms:

We are solving for the value of , which means we need to isolate the  to one side of the equation. We can subtract  from both sides:

Then divide both sides by  to solve for 

Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both  and  values. 

Now that we have the value of , we can plug that value into the  variable in one of our given equations and solve for 

Our point of intersection, and the solution to the two system of linear equations is 

There are a couple of ways to solve a system of linear equations: graphically and algebraically. In this lesson, we will review the two ways to solve a system of linear equations algebraically: substitution and elimination. 

Substitution can be used by solving one of the equations for either  or , and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the  form, and then set both equations equal to each other. 

Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable. 

For this problem, substitution makes the most sense because the first equation is already solved for a variable. We can substitute the expression that is equal to , into the  of our second equation:

Next, we need to distribute and combine like terms:

We are solving for the value of , which means we need to isolate the  to one side of the equation. We can subtract  from both sides:

Then divide both sides by  to solve for 

Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both  and  values. 

Now that we have the value of , we can plug that value into the  variable in one of our given equations and solve for 

Our point of intersection, and the solution to the two system of linear equations is 

There are a couple of ways to solve a system of linear equations: graphically and algebraically. In this lesson, we will review the two ways to solve a system of linear equations algebraically: substitution and elimination. 

Substitution can be used by solving one of the equations for either  or , and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the  form, and then set both equations equal to each other. 

Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable. 

For this problem, substitution makes the most sense because the first equation is already solved for a variable. We can substitute the expression that is equal to , into the  of our second equation:

Next, we need to distribute and combine like terms:

We are solving for the value of , which means we need to isolate the  to one side of the equation. We can subtract  from both sides:

Then divide both sides by  to solve for 

Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both  and  values. 

Now that we have the value of , we can plug that value into the  variable in one of our given equations and solve for 

Our point of intersection, and the solution to the two system of linear equations is 

There are a couple of ways to solve a system of linear equations: graphically and algebraically. In this lesson, we will review the two ways to solve a system of linear equations algebraically: substitution and elimination. 

Substitution can be used by solving one of the equations for either  or , and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the  form, and then set both equations equal to each other. 

Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable. 

For this problem, substitution makes the most sense because the first equation is already solved for a variable. We can substitute the expression that is equal to , into the  of our second equation:

Next, we need to distribute and combine like terms:

We are solving for the value of , which means we need to isolate the  to one side of the equation. We can subtract  from both sides:

Then divide both sides by  to solve for 

Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both  and  values. 

Now that we have the value of , we can plug that value into the  variable in one of our given equations and solve for 

Our point of intersection, and the solution to the two system of linear equations is 

There are a couple of ways to solve a system of linear equations: graphically and algebraically. In this lesson, we will review the two ways to solve a system of linear equations algebraically: substitution and elimination. 

Substitution can be used by solving one of the equations for either  or , and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the  form, and then set both equations equal to each other. 

Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable. 

For this problem, substitution makes the most sense because the first equation is already solved for a variable. We can substitute the expression that is equal to , into the  of our second equation:

Next, we need to distribute and combine like terms:

We are solving for the value of , which means we need to isolate the  to one side of the equation. We can subtract  from both sides:

Then divide both sides by  to solve for 

Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both  and  values. 

Now that we have the value of , we can plug that value into the  variable in one of our given equations and solve for 

Our point of intersection, and the solution to the two system of linear equations is 

There are a couple of ways to solve a system of linear equations: graphically and algebraically. In this lesson, we will review the two ways to solve a system of linear equations algebraically: substitution and elimination. 

Substitution can be used by solving one of the equations for either  or , and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the  form, and then set both equations equal to each other. 

Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable. 

For this problem, substitution makes the most sense because the first equation is already solved for a variable. We can substitute the expression that is equal to , into the  of our second equation:

Next, we need to distribute and combine like terms:

We are solving for the value of , which means we need to isolate the  to one side of the equation. We can subtract  from both sides:

Then divide both sides by  to solve for 

Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both  and  values. 

Now that we have the value of , we can plug that value into the  variable in one of our given equations and solve for 

Our point of intersection, and the solution to the two system of linear equations is 

There are a couple of ways to solve a system of linear equations: graphically and algebraically. In this lesson, we will review the two ways to solve a system of linear equations algebraically: substitution and elimination. 

Substitution can be used by solving one of the equations for either  or , and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the  form, and then set both equations equal to each other. 

Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable. 

For this problem, elimination makes the most sense because our  variables have the same coefficient. We can subtract our equations to cancel out the 

Next, we can divide both sides by  to solve for 

Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both  and  values. 

Now that we have the value of , we can plug that value into the  variable in one of our given equations and solve for 

We want to subtract  from both sides to isolate the 

Then divide both sides by  to solve for 

Our point of intersection, and the solution to the two system of linear equations is 

First, translate the problem into three equations. The statement, "Joule is three years older than twice Newton's age" is mathematically translated as

where  represents Joule's age and  is Newton's age.

The statement, "Newton is Toby's age younger than eleven years" is translated as

where  is Toby's age.

The third statement, "Toby is one year younger than Joule" is

.

So these are our three equations. To figure out the age of these dogs, first I will plug the third equation into the second equation. We get

Plug this equation into the first equation to get

Solve for . Add  to both sides

Divide both sides by 3

So Joules is 9 years old. Plug this value into the third equation to find Toby's age

Toby is 8 years old. Use this value to find Newton's age using the second equation

Now, we have the age of the following dogs:

Joule: 9 years

Newton: 3 years

Toby: 8 years

Since this is a long word problem, it might be easy to confuse the two behaviors and come up with the wrong answer. Let's avoid this problem by turning each behavior into a variable. If we call "sitting quietly"  and "completing assignments" , then we can easily construct a simple system of equations, 

and 

.

We can multiply the first equation by  to yield .

This allows us to cancel the  terms when we add the two equations together. We get , or .

A quick substitution tells us that . So, sitting quietly is worth 3 tokens and completing an assignment on time is worth 7.

Let  be the number of adult tickets sold. Then the number of child tickets sold is .

The amount of money raised from adult tickets is ; the amount of money raised from child tickets is . The sum of these money amounts is , so the amount of money raised can be defined by the following equation:

To find the number of adult tickets sold, solve for :

 adult tickets were sold.