Any time the product of consecutive numbers is , must be a one of those consecutive numbers, because if it is not, the product will be non-zero. This leaves us with four possibilities, depending on where is placed in the sequence.

As we can see, , , and are our numbers in question, meaning is our answer as the lowest number.

Note that it is possible to use algebra and set up a system of equations, but it's more time-consuming, which could hinder more than help in a standardized test setting.

You can subtract the second equation from the first equation to eliminate y:

7_x_ + y = 25 – 6_x_ + y = 23: 7_x_ – 6_x_ = x; y – y = 0; 25 – 23 = 2

x = 2

You could also solve one equation for y and substitute that value in for y in the other equation:

6_x_ + y = 23 → y = 23 – 6_x_.

7_x_ + y = 25 → 7_x_ + (23 – 6_x_) = 25 → x + 23 = 25 → x = 2

Any time the product of consecutive numbers is , must be a one of those consecutive numbers, because if it is not, the product will be non-zero. This leaves us with four possibilities, depending on where is placed in the sequence.

As we can see, , , and are our numbers in question, meaning is our answer as the lowest number.

Note that it is possible to use algebra and set up a system of equations, but it's more time-consuming, which could hinder more than help in a standardized test setting.

You can subtract the second equation from the first equation to eliminate y:

7_x_ + y = 25 – 6_x_ + y = 23: 7_x_ – 6_x_ = x; y – y = 0; 25 – 23 = 2

x = 2

You could also solve one equation for y and substitute that value in for y in the other equation:

6_x_ + y = 23 → y = 23 – 6_x_.

7_x_ + y = 25 → 7_x_ + (23 – 6_x_) = 25 → x + 23 = 25 → x = 2

If we use elimination to solve this system of equations, we can add the two equations together. This results in 0=0.

When elimination results in 0=0, that means that the two equations represent the same line. Therefore, there are an infinite number of solutions.

If we use elimination to solve this system of equations, we can add the two equations together. This results in 0=0.

When elimination results in 0=0, that means that the two equations represent the same line. Therefore, there are an infinite number of solutions.

In order to find this point, we must find the solution to the system of equations. we will use substitution, setting the two expressions for y equal to one another.

Then we plug this value back into either expression for y, giving us

So the point is (1, 6).

In order to find this point, we must find the solution to the system of equations. we will use substitution, setting the two expressions for y equal to one another.

Then we plug this value back into either expression for y, giving us

So the point is (1, 6).

You can solve this problem in a number of ways, but one way to solve it is by using substitution. You can begin to do that by solving for in the first equation:

Now, you can substitute in that value of into the second equation and solve for :

Let's consider this equation as adding a negative 3 rather than subtracting a 3 to make distributing easier:

Distribute the negative 3:

We can now combine like variables and solve for :

You can solve this problem in a number of ways, but one way to solve it is by using substitution. You can begin to do that by solving for in the first equation:

Now, you can substitute in that value of into the second equation and solve for :

Let's consider this equation as adding a negative 3 rather than subtracting a 3 to make distributing easier:

Distribute the negative 3:

We can now combine like variables and solve for :