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 .

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

 Divide both sides by .

 Rearrange the right side to match slope-intercept form.

Second equation:

 This equation is already in slope-intercept form.

Notice how the two equations are identical? This means the lines do not intersect at one point -- they intersect at all points, and are identical lines.

Let's use the method of substution to solve this. First we need to get one variable in terms of the other variable. Look at the 2nd equation and isolate x:

Moving y to the right side:

Now plug in this value of x into the first equation and solve for y:

Now plug y back into to either of the oringial two equations and solve for x:

Our answer is: 

This system of equations simplifies very nicely if one begins by isolating the variable q in the second equation. Isolating q in the second equation yields: . Now substitute this expression for q into the first equation. This results in: . Now distribute the one-third term outside the brackets to obtain: . This expression will simplify quite simply to: .

We can rewrite the first equation as:

If we substitute this new value for  into the second equation we get:

Simplify.

Combine like terms

Solve for 

Now substitute this value into either of the original equations:

While substitution is a plausible way to solve this problem, it does not provide the most effecient solution. Instead, begin by subtracting the first equation from the second equation in a manner similar to the "elimination" method used by many students.

This subtraction will result in 

.

Now, simply divide this result by 3 to obtain 

.

This result, indicates that the correct answer choice is 15. This problem is an excellent example of a time trap. The use of substitution would require use of fractions or time consuming conversions to decimals. If, however, the test taker identifies the route between the original two functions and the desired function of unkown value, then this problem becomes much quicker. 

First notice that in this system of equations, there are four unkown variables, but only three independent equations. Therefore, we know the system is indeterminate and so substitution/elimination will not work. The answer, however, is NOT "None of the above". The explanation is below:

Note that the third equation is extraneous information. Begin by look at the second equation, . This equation tells us that at least one of the variables (x, y, or z) must equal zero in order for the overall expression to equal zero. Now, look at the first equation . This equation indicates that NONE of the variables (r,z, nor x) are equal to zero, as the overall expression does NOT equal zero. Realize that "y" is the only variable that exists in the second expression, but does not exist in the first expression.

Therefore, we can definitively state that 

To find the solution we will multiply by some number to cancel out one variable and then solve for the variable that is left over. This is called the "elimination" method. You may choose to eliminate either variable, but it is prudent to choose the variable that is easier to cancel.

1) 

2) 

Multiply equation 1 by -2:

Add new equation 1 to equation 2:

The result is 

Solve for y:

Now solve for x by plugging in the value for y (we can choose either original equation):