The question is asking the student to solve the linear set of equations ultimately by isolating and .

There are a few ways a student could choose to answer this. 

One may see immediately  and realize to eliminate the result from the second equation one would subtract as follows.

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Other students may choose to subtract the first equation twice from the second equation then subsequently solve for y.

The point these two equations intersect is their solution. There are several ways to solve and one should choose the method that makes the problem easiest. Here one can easily eliminate the two 4x variables. Thus, this problem will be solved by elimination.

1) 

2) 

Multiply equation 1 by -1:

Add equation 1 to equation 2:

1a)  

2)    

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Now substitute for y in either of the original equations:

2) 

The solution set to this system of equations is a coordinate point whose x and y values would satisfy the two equations. This is the same place the two equations intersect. The elimination method will be used to solve for the solution set.

1) 

2) 

Multiply equation 1 by -2:

Add the new equation 1 to equation 2:

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Substitute this x value back into either original equation to solve for y:

First step is to solve one of the equations for one of the variables.

Choose the equation easiest to solve for one of the variables.

.             

Substitute  for  in the other equation and solve for .

Use distributive property.

Substitute  for  in either equation and solve for .

The solution is 

For elimination you need to get one variable by itself by cancelling the other out. In this equation this is best done by getting rid of . You can multiply whichever equation you would like to, but multiply it by  to get

then add the equations together

which, simplified, is

divied by  to get  

Then plug  back into any equation for the x value

Solve for  to get

The first step is to set the two equations equal to each other, since at the point of intersection, they will be. To do this more easily, convert each equation into slope-intercept form.

First equation:

 State equation

 Add  to both sides.

Second equation:

 State equation

 Divide both sides by .

 Subtract  from both sides.

Now, since each equation equals , the equations also equal each other (for the point of intersection). By solving for , therefore, we can 

 State equation.

 Subtract  from both sides.

 Subtract  from both sides.

 Divide both sides by .

So, the -coordinate of our intersection is . To find the -coordinate, plug this result back into one of the original equations.

 State your chosen equation.

 Substitute the value of .

So, the coordinates where the two lines intersect are .

This solution set is special, as one of our two lines is either horizontal or vertical. If the equation of a line in the coordinate plane contains only an  or a  variable, but not both, then the line is either horizontal (if only  is in the equation) or vertical (if only  is in the equation). This is good news, as solving for this intersection is much faster.

First, solve for  in our first equation.

 State the equation.

 Divide both sides by .

Now that we know , we simply substitute that value into our second equation.

 State the equation.

 Substitute the value of .

 Simplify.

 Subtract  (or ) from both sides.

 Divide both sides by  (or multiply both sides by ).

Thus, the intersection point of our two lines is at .

The first step is to set the two equations equal to each other, since at the point of intersection, they will be. To do this more easily, convert each equation into slope-intercept form.

First equation:

 Already in slope-intercept form

Second equation:

 State equation

 Add  to both sides.

 Divide both sides by .

At this point, note that both equations have idential slopes:  for both equations, but different -intercepts. Thus, the lines are parallel, and will never touch. We can stop here, but let's prove our theory with algebra by setting the equations equal to one another:

 Set your equations.

 Subtract  from both sides.

 No solution.

Thus, there is no solution to this equation, and the lines are parallel.

The first step is to set the two equations equal to each other, since at the point of intersection, they will be. To do this more easily, convert each equation into slope-intercept form.

First equation:

 State equation

 Add  to both sides.

 Divide both sides by .

Second equation:

 State equation

 Symmetric Property of Identity

Now, since each equation equals , the equations also equal each other (for the point of intersection). By solving for , therefore, we can 

 State equation.

 Add  to both sides.

 Subtract  from both sides.

 Divide both sides by  (or multiply both sides by ).

So, the -coordinate of our intersection is . To find the -coordinate, plug this result back into one of the original equations.

 State your chosen equation.

 Substitute the value of .

 Multiply.

 Subtract.

So, the coordinates where the two lines intersect are .

The first step is to set the two equations equal to each other, since at the point of intersection, they will be. To do this more easily, convert each equation into slope-intercept form.

First equation:

 State equation

 Multiply both sides by .

Second equation:

 State equation

 Subtract  from both sides.

 Divide both sides by .

 Rearrange (symmetric property of equality).

Now, since each equation equals , the equations also equal each other (for the point of intersection). By solving for , therefore, we can 

 State equation.

 Add  to both sides.

 Add  to both sides.

 Divide both sides by .

So, the -coordinate of our intersection is . To find the -coordinate, plug this result back into one of the original equations.

 State your chosen equation.

 Substitute the value of .

 Distribute.

 Simplify.

So, the coordinates where the two lines intersect are .