Textbook

*Systems* is a term used to describe a combination of two or more equations where the task is typically to solve for one or more of the variables. On the CLT, systems will typically involve two equations; in rare cases, on hard problems located near the end of the Quantitative Reasoning section, the test will present three equations and ask the test taker to solve for one of the variables. Systems questions often present the student with multiple ways of solving them, including substitution, elimination, and graphing. We’ll explore these below.

How many points of intersection do the following equations have?

$y=3x_{2}−7$

$y=2x+1$

A. 0

B. 1

C. 2

D. 3

This question is a gold mine of possibility; there are at least three ways to approach it, each of which touches on key concepts and skills for the CLT.

Exploring the systems of equations aspect of this problem first, we can observe the important detail that $y$ is isolated in both equations; by the transitive property, which tells us that two quantities equal to a third quantity must also be equal to each other, we knows we can set the right sides of the two equations equal to each other. Another way to say this is that we can *substitute* either right side for $y$ in the other equation. Since y is equal to both quantities on the right, either of those quantities can always be exchanged for $y$. This is the magic of the equals sign!

Setting the two right sides equal yields $3x_{2}−7=2x+1$. As we note in the quadratics lesson, we encounter here the important principle of *always setting quadratic expressions equal to zero*. We want to retain the positive coefficient in front of the $x_{2}$ terms, so we’ll subtract 2x from both sides and add 4 to both sides. This gives us $3x_{2}−8x−3=0$.

At this point, we have two choices: factoring or quadratic formula. If you are comfortable factoring a quadratic with a leading coefficient (there are multiple methods), try it now. Here’s the answer: $(3x+1)(x−3)$. It is clear from this that there are two distinct solutions: $−⅓$ and $3$. The answer is **C**.

Because this sort of factoring can be challenging and does not occur very frequently on the CLT, we recommend honing your skills in using the quadratic formula so you always have the option to solve this way. Using the quadratic formula, we create a numerator of $−(−8)±−8)_{2}−(4)(3)(−3 $ and a denominator of $(2)(3)=6$.

That’s kind of a mess, but there’s good news: we don’t need to solve all the way! To identify the number of solutions when a quadratic is involved, we’ll use the *discriminant*: the portion of the quadratic formula under the radical, or $b_{2}−4ac$. If that solution is positive, there are two solutions to the equation; if it’s zero, then there is one solution; if negative, there are no real solutions (the solutions are imaginary or, put another way, the parabola does not intersect with the x-axis on the real number xy-coordinate plane). In this case, we know we have a positive number when $−8$ is squared; we also know that we’ll be adding to that number because the two negatives will cancel out. The value is $64+36=100$, but again, we don’t need the actual value; we just need to know that the discriminant is positive. We have confirmed answer **C**, 2 solutions.

There is a takeaway shortcut from this process: *if a quadratic has a negative c term, that quadratic will always have two positive roots.* Why is that? Think about the nature of $b_{2}−4ac$: we know that $b_{2}$ is greater than or equal to zero and that $−4ac$ will be *positive* if the $c$ is negative. So we’re adding a positive number to a nonnegative number. The result must always be positive! *Note*: this inference only holds true with a positive leading coefficient, so make sure that when you set the quadratic equal to zero, you create a positive $x_{2}$ term. If that term is negative, you simply need to multiply the entire equation by $−1$.

We have encountered two of three plausible ways to solve this problem: after substitution, we had factoring or the quadratic formula. If you are confident in your graphing skills, you may want to try a graphing approach to this problem. Graphing on the CLT can be of limited value because it must be done by hand, without a calculator. But this problem doesn’t necessarily require precise graphing; it only requires visualizing the line and the parabola and considering their intersection. If you remember how the basic “parent” form of the parabola, $y=x_{2}$, is graphed, you’ll know that it intersects the origin and rises on both sides of the origin. Our parabola here is translated *down* seven units because of the $−7$. The $3$ on the front stretches the parabola on the y-axis, making it “skinnier”. Meanwhile, the line has a positive y-intercept at (0,1). If you can draw this well or even visualize it in your mind, you can see that we don’t even need to consider the line’s slope because it *must* intersect the parabola twice, once on the right of the y-intercept and once on the left. We have graphing-based confirmation: two points of intersection.

One final note: graph paper *is* included among the permitted options for scratch paper on the CLT, so we recommend practicing with graph paper to get used to using it in cases like this question.

Though most CLT systems involve two equations, you may encounter a system of three equations late in the CLT Quantitative Reasoning section. Typically, one of two pathways will be available. 1) There may be an opportunity to investigate the possibilities using “guess and check.” For example, if one of the equations is $x_{y}=16$ and you know the variables represent integers, there are only three possibilities: either $x$ is $2$ and $y$ is $4$, or $x$ is $4$ and $y$ is $2$, or $x$ is $16$ and $y$ is $1$. You can plug these possibilities into the other equations to see what works. 2) One of the equations will only employ two of the three variables unknown - say, $a$ and $b$, but not $c$. In this case, you should use elimination with the other two equations to reduce them to another equation involving $a$ and $b$. Then you can treat what remains as a normal system of two equations.

The other variation you may encounter is a system of *inequalities* rather than equations. Some of the same tools used for systems of equations will apply with inequalities as long you keep the following caveats in mind:

- Elimination will work with systems of inequalities, but substitution will not.
- You can only perform elimination if the two inequality symbols are pointing in the same direction (don’t try it with one “less than” and one “greater than”).
- Eliminate
*only*by adding, not by subtracting. (Subtracting affects the second inequality as if you’re multiplying by $−1$, which would change the direction of the inequality. That’s some quicksand you want to avoid!)

With elimination, you may want to note the following on flashcards:

- If the same variable in each equation has the same sign, you should
*subtract*the two equations (**S**ame =**S**ubtract). - If the variables have coefficients that are equal except they have opposite signs, you should
*add*the equations to eliminate a variable.

Which of the following coordinate points is a solution to the system of equations below?

$3x−y=4$

$2x+y=1$

A. (0,-1)

B. (1,-1)

C. (2,1)

D. (3,4)

(spoiler)

The answer is **B**. This is an excellent chance to practice *elimination*, because the $y$ coefficients are equal but opposite. By simply adding the equations as they stand, we will eliminate the $y$ variables and be left with $5x=5$. So $x=1$. We could plug 1 back into either equation and find that $y=−1$, but use the UnCLES and be smart: if $x−1$, there’s only one possible answer already!

Which of the following is a solution to the system of inequalities below?

$x+2y<5$

$3x−y>4$

A. (0,5)

B. (1,2)

C. (2,1)

D. (0,2)

(spoiler)

The answer is **C**. Systems of inequalities are a little different than systems of equations (see discussion under *Variations* in this lesson). We could work to switch the signs of inequality, which would involve multiplying through that inequality by $−1$. But even then, the coefficients would not be the same either for $x$ or for $y$. Much better to simply plug in the values.

Choice A doesn’t work for the first inequality (10 is not less than 5), so we can eliminate it without checking the second inequality. Choice B also doesn’t work for the first one (tricky, but $5$ is not less than $5$). Choice C gives us $4<5$ for the first inequality, so we proceed to the second. That yields 5>4, so both are true; this is the answer. Choice D works for the first inequality ($4<5$), but not for the second ($−2$ is not greater than $4$).

Which of the following coordinate points is a solution to both the equations below?

$y=3x/4$

$(−1/8)x_{2}=−2+y$

A. (-8,-6)

B. (-2,-3/2)

C. (0,0)

D. (4,4)

(spoiler)

The answer is **A**. We certainly have the option of backsolving (plugging in the answers) on this question, and that will be the fastest approach for many students. But let’s explore the textbook approach here. Since y is isolated in the first equation, let’s substitute its equivalent value, $3x/4$, into the second equation. Then we’ll multiply through by $−8$ to get rid of the negative fractional coefficient before $x_{2}$. We now have $x_{2}=16−6x$. Bringing all the terms to one side leaves $x_{2}+6x−16=0$, and that quadratic factors so that we end up with $(x+8)(x−2)=0$. So $x=−8$ and $x=2$. But only the first of these is present in the answer choices. If we want to make sure, we can plug answer choice A back into both equations, using $−8$ for $x$, and in both cases $y=−6$.

If we do choose to backsolve, it would be sense to start with answer choice $C$, since $0$ is a simple number to plug in. $0$ works for the first equation but not the second. We could then move to choice $D$ because it has two positive integers, but that choice works for neither equation. We would likely try $A$ next to avoid dealing with the fraction in choice $B$, and careful plugging in of the numbers in choice $A$ confirms our answer.

How many solutions does the following system of equations have in the real number system?

$y=−x_{2}+4x−3$

$y=2x−1$

A. 0

B. 1

C. 2

D. 3

(spoiler)

The answer is **A**. This problem calls for finding the quadratic’s discriminant once again, but with a different result from our approach question. We start as usual by setting the right sides equal to each other; however, we should eliminate the negative in front of the $x_{2}$ at the earliest opportunity, so let’s multiply the resulting equation by $−1$. We now have $x_{2}−4x+3=−2x+1$. Moving the terms to one side, we get $x_{2}−2x+2$. That doesn’t look like it will factor, so let’s employ the quadratic formula or, more specifically, the discriminant. $b_{2}−4ac$ here leaves $4−(4)(2)$, which is $−4$. A negative result for the discriminant means zero solutions in the real number system.

Which of the following is the x-coordinate of a point (x,y,z) that would satisfy all three equations below?

$x−7+z=y_{3}$

$2x=z−2$

$y_{2}=4$

A. -8

B. -4

C. -1

D. 0

(spoiler)

The answer is **C**. This question looks intimidating, but it helps to start with the most accessible equation. The last equation might look extremely simple, but remember that taking the square root of both sides requires adding a “plus/minus” symbol to show that, in this case, $y$ could equal either $2$ or $−2$. The best approach is to plug in both of these options and see which works. If $y=2$, then the first equation is now $x−7+z=8$, which simplifies to $x+z=15$. We can combine that with the other equation containing $x$ and $z$; if you follow that process, you’ll find that x does not come out as a whole number and therefore doesn’t match the answer choices.

By elimination, then, we now try $y=−2$. That makes the first equation $x−y+z=−8$, or $x+z=1$. Combined with the equation $2x=z−2$, we find that $x=−1$.

Although it might be time-consuming, you can also do this problem in reverse. To abbreviate the process, we’ll go right to plugging in the right answer. If $x=−1$, then by the second equation $z=0$. Using those two values in the first equation yields $y=−2$, which makes the last equation true.

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