Systems of Equations and Inequalities

The answer is $(1,-1)$. There are numerous ways to solve this problem. If you are adept with Desmos, it may be relatively fast to graph the equations and locate the point of intersection. However, the problem also sets up quite nicely for elimination; simply adding the equations eliminates the y variables and gives us $5x=5$. That tells us that $x=1$, which, if we use the UnCLES method and consider the answers, tells us the answer already.

Those who like plugging in the answers could also try that approach, though it may not be the fastest in this case. The least preferred approach here is substitution, since no variables are currently isolated; the substitution method would entail several steps to arrive at the solution.

Note: the “plug and chug” method works well here because all the answers are integers. The SAT may include “messier” numbers, but you can still backsolve quite effectively in those cases with the help of the calculator.

The answer is $(2,1)$. Systems of inequalities are a little different than systems of equations (see discussion under Variations in this lesson). Elimination can be used to solve systems of inequalities, but only if you are adding the inequalities (do not subtract!). To do that, 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.

$(0.5)$ doesn’t work for the first inequality ($10$ is not less than $5$), so we can eliminate it without checking the second inequality. $(1,2)$ also doesn’t work for the first one (tricky, but $5$ is not less than $5$). $(2,1)$ 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. $(0,2)$ works for the first inequality ($4<5$), but not for the second ($-2$ is not greater than $4$).

The answer is $-5$. Desmos is a great approach to this question; the more you practice it, the more comfortable you will feel, and the multiple-choice answers tell you that the lines will definitely cross at integer coordinates (which makes them easier to read, since they’re right on the gridlines). The point of intersection is $(1,-5)$; just make sure you answer for the y-value of $-5$, since the $x$-value of $1$ is there as the trap answer!

For those who prefer a non-graphing approach, elimination is far superior to substitution in this case. To solve by elimination, we look for factor-multiple relationships between the $y$-coefficients or the $x$-coefficients. The y-coefficients don’t work because $3$ does not divide evenly into $10$; since $4$ divides evenly into $12$, let’s multiply the first equation by $3$ to equalize the $x$-coefficients. We can then subtract the equations as follows:

$$\begin{array}{rl}12x-9y& =57\\ -(12x-10y& =62)\\ y& =-5\end{array}$$

If you prefer to avoid having to eliminate by subtraction, multiply the first equation by -3 instead of 3. That will make the signs for the x-coefficients opposite to each other, and those terms will cancel when you add the equations together.

The answer is $1$. This is an interesting problem whose outcome is often misunderstood. Fortunately, Desmos breaks through the confusion; simply graphing these equations shows that there is one point of intersection: at the y-intercept of $7$. If you recognize slope-intercept form, that should make sense; both equations have a $y$-intercept of $7$. Further, since they have different slopes, they should cross exactly once.

If you don’t use Desmos here, a good approach is to set the right sides of both equations equal to each other, since they’re both each to $y$. (The transitive property tells us that two expressions equal to the same value must also be equal to each other.) This is where some confusion can come in. If we start with $3x+7=2x+7$ and subtract $7$ from both sides, we are left with $3x=2x$. Many students will think this means that there are no solutions, but there is in fact one solution that works: zero! Subtracting $2x$ from both sides of this most recent equation, after all, yields, $x=0$. If there is only one value that works for $x$, there must also be only one answer that works for y, and we need not discover what that is (spoiler: it’s $7$!) in order to answer the question.

The answer is $\frac{1}{3}$. This question looks intimidating because there are multiple unknowns. Also, having three unknowns makes it difficult to solve via Desmos (unless you plug in a number for one of the unknowns and call the other two $x$ and $y$). But, as is often the case, hidden similarities simplify the process.

Examination of both equations reveals that, in each equation, the a on both sides can be immediately canceled. In the first equation is can be canceled through division; in the second, through multiplication. This simplifies things considerably, leaving us with $\frac{1}{b}=27$ and $\frac{b}{n}=\frac{1}{9}$.

Since the first equation has only one unknown now, we should begin there, solving for $b$. Multiplying both sides by $b$ and dividing both by $27$, we get $b=\frac{1}{27}$. (Notice that if the variable starts out in the denominator of one side, it ends up “switching places” with the quantity in the numerator on the other side.)

Now we can plug in $\frac{1}{27}$ for $b$ in the second equation. We must be careful to locate the $27$ in the denominator, to be multiplied by the $n$ that is “already there”. The result is $\frac{1}{27n}=\frac{1}{9}$. Cross-multiplying shows us that $27n=9$. Dividing both sides by $7$, we see that $n=\frac{1}{3}$.