When you're facing systems of equations, the SAT often provides you a shortcut if you recognize it. When a systems of equation question asks you to solve for a combination of variables (e.g. $\displaystyle x+y$) and not just a single variable, there's often a faster way to solve directly for the combination.

Here, recognize that if you simply add the two equations together - much like using the "Elimination Method" but without actually trying to eliminate a variable - you can get $\displaystyle x$ and $\displaystyle y$ together with the same coefficient.

$\displaystyle 7x+y=25$

$\displaystyle x+7y=31$

sums to:

$\displaystyle 8x+8y=56$

If you then divide both sides by 8, you can get to exactly the answer they're looking for:

$\displaystyle x+y=7$

This system of equations provides you an excellent opportunity to use the Elimination Method to isolate a single variable. With the two equations provided, you already have a negative $\displaystyle x$ term (in the second equation) and a positive $\displaystyle x$ term (in the first).  If you multiply the second equation by 2, you can get the coefficients the same and sum the equations to arrive at a single variable, $\displaystyle y$. First, multiply the entire second equation by 2:

$\displaystyle 2(y-x=3) = 2y-2x=6$

Then stack the updated equations and sum them:

$\displaystyle y + 2x=9$

$\displaystyle 2y-2x=6$

Gives you:

$\displaystyle 3y=15$

This means that $\displaystyle y=5$ but remember to always double check that you've solved for the proper variable. The question wants to know $\displaystyle x$ so you can plug $\displaystyle y=5$ back into either equation to solve. Using the original second equation:

$\displaystyle 5-x=3$

So 

$\displaystyle x=2$

This problem asks you to solve for $\displaystyle xy$, and conveniently provides you with a single-variable equation that allows you to solve directly for $\displaystyle x$ so that you're halfway home already:

$\displaystyle 3x=12$, so if you divide both sides by $\displaystyle 3$ you'll find that $\displaystyle x=4$.

You can then plug in $\displaystyle x=4$ to the first equation to get:

$\displaystyle 10y-4(4)=34$

Adding $\displaystyle 16$ to both sides gives you:

$\displaystyle 10y=50$

And then divide both sides by $\displaystyle 10$ to get $\displaystyle y=5$.

You now have your values for $\displaystyle x$ and $\displaystyle y$ so you can multiply them to get $\displaystyle xy$:

$\displaystyle 4\times 5=20$, making $\displaystyle 20$ the correct answer.

This problem affords you a quick opportunity to use the Elimination Method. The first equation has a positive $\displaystyle x$ term and the second has a negative $\displaystyle x$ term, meaning that if you sum the two equations you will eliminate the $\displaystyle x$ terms and be left with a single variable, $\displaystyle y$.

The two equations sum to:

$\displaystyle -y=-4$

And dividing both sides by $\displaystyle -1$ allows you to determine that $\displaystyle y=4$.

Note that the question asks for $\displaystyle x$, not $\displaystyle y$, so you need to plug $\displaystyle y=4$ back in to one of the two equations to solve for $\displaystyle x$. Using the first equation, you have:

$\displaystyle x-3(4)=1$

So $\displaystyle x-12=1$ meaning that $\displaystyle x=13$.

One way to look at a system of linear equations is that the solution to that system is the point at which the graphs of the lines intersect at the same $\displaystyle (x,y)$ point. So a system of linear equations WILL NOT have any solutions if the lines never meet; in other words, if the lines are parallel with different y-intercepts.

The "by the book" method to determine if lines are parallel is to put each into slope-intercept form, $\displaystyle y=mx+b$, and then see if the slopes $\displaystyle (m)$ are the same. For the first line, that's:

$\displaystyle y=-2x+8$

So for the second line, you would need to find the $\displaystyle a$ value that makes the slope equal to -2. To get closer to slope-intercept form of the second equation you can start at:

$\displaystyle ay=6x+10$

And then divide both sides by $\displaystyle a$ to isolate the $\displaystyle y$ term:

$\displaystyle y=\frac{6x}{a}+\frac{10}{a}$

If $\displaystyle a=-3$ you've matched the slope of $\displaystyle -2$, making $\displaystyle -3$ the correct answer.

Of course, there's a shortcut to this. If you recognize that between the two equations, the scale factor from first to second is that $\displaystyle x$ is multiplied by 3, you can choose a value for $\displaystyle a$ that provides the same effect for the $\displaystyle y$ term. Since $\displaystyle y$ is multiplied by $\displaystyle -a$ in the second equation, you can say that:

$\displaystyle -a=3$

So $\displaystyle a=-3$.

For this problem, there are reasons to choose either the Elimination Method or the Substitution Method to solve the system. The question asks for the value of $\displaystyle a$ and the first equation gives you a great opportunity to substitute for $\displaystyle b$ in terms of $\displaystyle a$ and use the Substitution Method.  Since $\displaystyle 2a+b=0$, you can conclude that:

$\displaystyle b=-2a$

And then substitute $\displaystyle -2a$ where $\displaystyle b$ appears in the second equation:

$\displaystyle 7a+3(-2a)=-2$

Distribute the multiplication across parentheses to get:

$\displaystyle 7a-6a=-2$

And then solve:

$\displaystyle a=-2$

Of course, you could also use the Elimination Method. If you multiply the first equation by -3, you'd get:

$\displaystyle -6a-3b=0$, which you can stack with the second equation and sum:

$\displaystyle 7a+3b=-2$

When you sum, the $\displaystyle b$ terms cancel leaving you with the answer:

$\displaystyle a=-2$

This system of equation gives you an opportunity to use the Elimination Method to quickly eliminate the variable $\displaystyle x$ and then use a single-variable equation to solve for $\displaystyle y$. If you multiply the first equation by 3, you will then have a $\displaystyle +3x$ term in the first equation and a $\displaystyle -3x$ term in the second, so adding the two equations will eliminate the $\displaystyle x$.

Once you've multiplied the first equation by 3, you'll sum the equations:

$\displaystyle 3x+\frac{3y}{2}=27$

$\displaystyle -3x+5y=-1$

Adding these together gives you:

$\displaystyle \frac{13y}{2}=26$

And then you can solve for $\displaystyle y$ by multiplying each side of the equation by $\displaystyle \frac{2}{13}$ to get $\displaystyle y=4$.

Note that you should always double check that you've solved for the right variable (or combination of variables) for the question. This question does ask for $\displaystyle y$ so your answer is $\displaystyle 4$.

You can subtract the second equation from the first equation to eliminate $\displaystyle y$:

$\displaystyle 7x+y=25-6x+y=23:\, 7x-6x=x;\, y-y=0;\, 25-23=2$

$\displaystyle x=2$

You could also solve one equation for $\displaystyle y$ and substitute that value in for $\displaystyle y$ in the other equation:

$\displaystyle 6x+y=23\rightarrow y=23-6x$

$\displaystyle 7x+y=25\rightarrow 7x+(23-6x))=25\rightarrow x+23=25\rightarrow x=2$

Whenever you have a chance to solve a system of equations using the Elimination Method, it is generally your fastest option. Here if you multiply the first equation by 2, you can then subtract the two equations:

$\displaystyle x(x-3y=-5)$ gives you a new first equation of $\displaystyle 2x-6y=-10$.  When you then stack and subtract:

$\displaystyle 2x-6y=-10$

$\displaystyle 2x+5y=12$

You're left with:

$\displaystyle -11y=-22$

So $\displaystyle y=2$.

Then just plug $\displaystyle y=2$ back in to one of the two equations and you can solve for $\displaystyle x$. If you use the first, you'll have:

$\displaystyle x-3(2)=-5$

$\displaystyle x-6=-5$

So $\displaystyle x=1$.  This means that the ordered pair is $\displaystyle (1,2)$.

This problem provides you with a quick opportunity to use the Elimination Method. Because there is an $\displaystyle x$ in the first equation and $\displaystyle a-x$ in the second, if you add the two equations you can eliminate the $\displaystyle x$ terms and solve straight for $\displaystyle y$:

$\displaystyle x+3y=30$

$\displaystyle 2y-x=5$

$\displaystyle 5y=35$

This means that $\displaystyle y=7$.  

Then you can plug $\displaystyle y$ into either of the equations and you'll solve for $\displaystyle x$. Using the first equation, that would be:

$\displaystyle x+3(7)=30$

$\displaystyle x+21=30$

So $\displaystyle x=9$

This means that the correct ordered pair is $\displaystyle (9,7)$.