Linear Equations

The answer is C. This is a straightforward “plug in” (substitution) question; we just need to be careful for which variable we substitute the $10$. The question says that it’s the $y$ variable that’s $10$, so can rewrite the equation as $10=5x-15$. We then solve as follows:

$$\begin{array}{rl}10& =5x-15\\ 25& =5x\\ 5& =x\end{array}$$

We will frequently remind you, after showing the solution to a problem, that you can always plug in the answers should you choose to do so. However, for a problem with this few algebraic steps, most students will find it faster and more reliable to simply solve the equation.

The answer is B. The UnCLES method reminds us to note key terms, and the term “parallel” is our important starting point for this question. It tells us that line $s$ has the same slope as line $t$. We can use the helpful formula $-A/B$ to find the slope of line $t$; its slope, and therefore also the slope of line $s$, is $-\frac{1}{2}$. The next step involves using the ever-helpful slope-intercept form: $y=mx+b$. Whenever we have the slope of a line and a point on the line, we can plug in both to slope-intercept form, leaving the only unknown as b. Remember that the $x$-intercept of $4$ is the point $(4,0)$, we have the following after substituting the slope for $m$ and the point’s coordinates for $x$ and $y$:

$$0=(-\frac{1}{2})(4)+b$$

Simplifying and solving gives us:

$$\begin{array}{rl}0& =-2+b\\ 2& =b\end{array}$$

If you’re wondering about alternate approaches to this question, you might consider graphing if you’re confident graphing lines based on one point and the slope. You could graph this line and discover that it crosses the $y$-axis at $2$. The challenge is to graph the accurately enough on your scratch paper to ensure that your answer is correct. If you plan to solve some linear equation questions on test day, we recommend you begin practicing with graph paper now. The CLT permits the use of graph paper for your scratch paper; you may also use lined paper and blank paper.

The answer is C. Don’t confuse parallel lines with perpendicular lines; parallel lines would have the same slope as the original line (and you can bet they’ll include a trap answer to that effect)! For this question, we’re looking for the negative reciprocal of the original line. So what’s the slope of the original line? Applying the slope formula, we get:

$$\begin{array}{rl}\text{slope}& =(4-(-5))/(-2-1)\\ \text{slope}& =(9)/(-3)\\ \text{slope}& =-3\end{array}$$

The negative reciprocal of $-3$ is $\frac{1}{3}.$

Note that we don’t have to take further steps, such as writing slope-intercept form. That would only be necessary if we were asked for the $y$-intercept, but here we only need the slope.

The answer is A. A word problem like this calls for the slope formula, but we need to set it up carefully. The vertical drop, given as $75$, corresponds to rise in the $\frac{rise}{run}$ formula. This means that, in the proportion, $75$ must be in the numerator. In addition, we can avoid later confusion about the sign of the result by noting that $75$ should be negative since there is a vertical drop. So the problem unfolds like this:

$$\begin{gathered} \frac{-1}{4}=\frac{-75}{x} \\ -x=-300 \\ x=300 \end{gathered}$$

In the steps above, we placed the negative sign in front of the numerator to make clear that the rise or vertical drop is really what’s negative.

Finally, note that answer choice $D$ results from setting up the proportion upside down, with the rise in the numerator and the run in the denominator.

The answer is B. As shown under Flashcard Fodder, a reflection negates the opposite variable from the axis over which it is reflected. Since this point is reflected over the $y$-axis, we negate the $x$-coordinate, turning it positive. The new point is $(2,6)$. What about the rotation? If you need to visualize this, graph the point $(2,6)$ and draw a circle going through it with its center at the origin. The new point should travel exactly halfway around that circle. Accordingly, the rotation will negate both coordinates, so that our point is now at $(-2,-6)$. Now we can make the point we just identified interact with the other given point using the slope formula. We’ve laid that out already twice in this lesson; try it on your own and see what you get. You should find the slope connecting the two lines is $-\frac{1}{2}$. We don’t need to concern ourselves with the $y$-intercept, because only one answer matches the slope we’ve discovered. (However, checking the $y$-intercept for answer B to make sure it fits our data would be an excellent way to make sure we haven’t made a mistake along the way. Problems with many steps should always be double-checked if time permits!

The answer is C. We first need to find the intersection point of lines $j$ and $k$. If both had been in slope-intercept form, we would have been able to set the $mx+b$ portions equal to each other, but one of these equations is in standard form. Since $y$ is isolated in the second equation, we can plug in the right side of that equation for $y$ so that we end up with $3x-7(-\frac{1}{7}x+5)=-7$. Simplifying gives us $3x+1x-35=-7$, then $4x=28$, then $x=7$. Plugging $7$ in for $x$ in either equation, we discover that $y$ is $4$. With this point in mind, we can find the equation for Line $l$ starting with $(7,4)$ and the new point $(-11,0)$. Using the slope formula, we find $\frac{4-0}{7-(-11)}$, which comes to $\frac{4}{18}$, which reduces to $\frac{2}{9}$.