Systems of Equations - Pre-Calculus

We can start by rearranging each equation so that y is on one side:

Now that both equations are equal to y, we can set the right sides equal to each other and solve for x:

Our last step is to plug these values of x into either equation to find the y values of our solutions:

So our solutions are the folloiwing two points:

Our first step is to rearrange each equation so that the left side is just y:

Now that both equations are equal to y, we can see that the right sides of each equation are equal to each other, so we set this up below and solve for x:

Our last step is to plug these values of x into either equation to solve for the y values of our solutions:

So the solutions to the system are the following points:

We can solve this equation by using substitution since the bottom equation is already solved for . Substituting the bottom equation into the top we get:

We then solve the equation for our values:

Finally, we substitute our values into the bottom equation to get our values:

Our different solutions are then:

Our first step is to solve the bottom equation for

We can now substitute it into the top equation:

and solve for our values:

our values are then:

We can now plug our values into the bottom equation we had solved for and arrive at our solutions:

So, our solutions are

We can substitute the top equation into the bottom:

and solve for values:

Now that we have our values we can plug it into the top equation and find our values

So, our values are

The first step is to solve the bottom equation for :

since our question specifies for we just focus on

We now substitute this equation into the top equation:

we can now plug in our x-values into the bottom equation to find our y-values:

The solutions are then:

Our first step is to solve the bottom equation for :

so we can substitute it into the top equation:

Now we can plug in our y-values into the bottom equation to find our x-values:

Remember we cannot take a square root of a negative number without getting an imaginary number. As such, we'll just focus on the values.

Our solution is then:

In order to solve a system of linear equations, we must start by solving one of the equations for a single variable:

We can now substitute this value for y into the other equation and solve for x:

Our last step is to plug this value of x into either equation to find y:

We first move to the left side of the equation:

Subtract the bottom equation from the top one:

Left Side:

Right Side:

So

So dividing by a -1 we get our result.

For any system of linear equations, we can start by solving one equation for one of the variables, and then plug its value into the other equation. In this system, however, we can see that both equations are equal to y, so we can set them equal to each other:

Now we can plug this value for x back into either equation to solve for y:

So the solutions to the system, where the lines intersect, is at the following point:

In order to solve a system of linear equations, we can either solve one equation for one of the variables, and then substitute its value into the other equation, or we can solve both equations for the same variable so that we can set them equal to each other. Let's solve both equations for y so that we can set them equal to each other:

Now we just plug our value for x back into either equation to find y:

So the solution to the system is the point:

Start from equation 3 because it has the least number of variables. We see directly that .

Back substitute into the equation with the next fewest variables, equation 2. Then,

. Solving for , we get

or .

Then back substitute our and into equation 1 to get

.

Solving for x,

.

So our solution to the system is

We can solve the system using elimination. We can eliminate our by multiplying the top equation by :

and then adding it to the bottom equation:

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We can now plug in our y-value into the top equation and solve for our x-value:

Our solution is then

We can solve the system using substitution since the bottom equation is already solved for :

Now we can plug in our value into the bottom equation to find our x-value:

So our solution is

We can solve the system using elimination. We can eliminate the x terms by multiplying the bottom equation by :

and now add it to the top equation:

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We plug in our y-value into the bottom equation to get our x-value:

Our solution is then

We can solve this system using either substitution or elimination. We'll eliminate them here.

Note: If you wanted to do substitution, we can do it by substituting the top equation into the bottom for .

We'll rearrange the bottom equation to have both y-values aligned and then add the equations:

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Now that we have our x-value, we can find our y-value:

Our answer is then

There are many ways to solve this system of equations. The following is just one way to reach the answer.

Add the two together, to elimnate the y variable. Solve for x and then plug it back in to the first equation to solve for y.

To solve this, let's first try to eliminate x. We can do this by adding the two equations:

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Which implies

We can now solve for x by plugging 2 in for y in either equation.

Thus we have the answer

Let's solve this equation by eliminating the variable x by adding a multiple of the second equation to the first.

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Now let's combine those y values and solve for y.

Now all we have to do is plug that in for y in either original equation to solve for x.

Thus this yields the intersection point

Because one of the variables, z, has already been isolated, let's use the substitution method to solve this system of equations. We know z = 1, so let's plug that into the middle equation to solve for y:

Now that we have found y, let's solve for x by plugging both y and z into the top equation:

Thus we have found that the point of intersection would be