[fs-toc-h2]Systems of Linear Equations

‍

With the addition of Desmos, solving systems has become much easier. Though, it is still important to understand how to solve systems algebraically. If you haven't learned how to, or need a reminder, watch the video below.

When given two systems of equations and asked to find the solution, you can use Desmos with no computations. Refer to the example below:

‍

The intersection of both lines will be the \((x, y)\) solution to the system.

‍

After graphing both equations, our solution set is \((4, 6)\), meaning that 6 is the value of \(y\).

‍

Choice B is the correct answer.

‍

[fs-toc-h3]Solving for Expressions

‍

Sometimes, graphing isn't the quickest way though. Remember, while getting the answer correct is important, you have to get that answer as fast as possible. On module 2 especially, you need as much time as you can get. Refer to the question below and try to solve it as quickly as possible.

While you could graph both equations and then add the solutions, it could be much quicker by adding both equations.

‍

In any system of equations, you're able to add both equations together by combining like terms.

‍

On the left side of the equations we can add all like terms, like so:

‍

\[ 5x + (-4x + y) = x + y \]

‍

The same thing is true when adding on the right side of the equations:

‍

\[ 15 + (-2) = 13 \]

‍

Putting both sides of our equation together, our final equation becomes:

‍

\[ x + y = 13 \]

‍

Choice C is the correct answer

‍

After some practice, you’ll be able to identify these things instantly, which means that a question like this could be solved in under 10 seconds.

Again, here you could graph both equations and plug them back into \(5x + 5y\), but you probably shouldn't. Remember, you’re able to add the top and bottom equations. \(2x + 3x = 5x\), and \(3y + 2y = 5y\), so after adding the left side of the equation you end up with the expression \(5x + 5y\), exactly what we’re solving for.

‍

To figure out what that equals, simply add the right side. \(1200 + 1300 = 2500\), making 2500 the final answer.

‍

‍

[fs-toc-h2]Number of Solutions

‍

In a system of two equations, there are three outcomes when it comes to solutions: one solution, no solution, and infinite solutions. It is VITAL to know what these mean before your test as it is one of the most tested subtopics within systems.

‍

[fs-toc-h3]One Solution

‍

A linear system having one solution means that the lines of both equations intersect at exactly one point. Whenever you’re given a question in which you have to solve for \((x,y)\, the system has one solution.

‍

‍

[fs-toc-h3]No Solution

‍

No solutions (and infinite solution) questions are where things kinda get messy, and constants get mixed in. Do not worry though, most of these questions are usually quite simple and just require you to understand the vocabulary.

‍

IMPORTANT NOTE: A constant is simply an unvarying number that isn't directly given to us, a number that cannot change.

‍

In most cases, we’ll be asked to solve for the value of these constants, which may seem much more familiar.

‍

You can basically think of constants as placeholders for numbers.

‍

There is no reason to get scared or freeze up when you see the word “constant,” just understand what the question is asking and begin solving for an answer.

‍

Back to no solutions. If a system has zero solutions, then their lines will never intersect: no values for \(x\) and \(y\) will satisfy both equations.

‍

The only way for this to happen is if the slopes are equivalent and the y-intercepts are different. In other words, a system has no solutions if the lines are parallel. Parallel lines NEVER meet.

‍

‍

Notice that both equations within the system have the same slope (2), making both lines parallel. These equations will never intersect, meaning that there is no solution. Let's look at an example testing this topic.

At first glance, the question may seem intimidating: 2 variables, 2 equations, a constant, a fraction, and way too many numbers. But with every single SAT question, you have to go step by step.

‍

Whenever given a system of linear equations (or any linear equation), always convert them into slope-intercept form \[ y = mx + b \]. In other words, get \(y\) by itself on one side, and everything on the other. We’re going to want to do this for both equations as we need to compare their slopes later on.

‍

\[ 48x - 64y = 48y + 24 \]

\[ 48x - 24 = 112y \]

\[ 37x - 314 = y \]

‍

Our final slope-intercept form of our top equation is then \[ y = \frac{37}{48}x - \frac{314}{112} \], so our slope is \(\frac{37}{48}\).

‍

Repeating the same process for our second equation (by dividing both sides by \(r\) to isolate \(y\)), you get a final equation of \[ y = -\frac{12}{r}x + \frac{1}{8r} \], so our slope is \(-\frac{12}{r}\).

‍

Going back to the question, it asks us for what value of \(r\) does the equation have no solutions. Remember, for a system to have no solutions both equations must share the same slope/be parallel.

‍

Setting both slopes equal and solving by cross multiplication will then give us our value for \(r\):

‍

\[ -\frac{12}{r} = \frac{37}{48} \]

\[ -12 \cdot 48 = 37r \]

\[ -576 = 37r \]

\[ r = -\frac{576}{37} \]

‍

If the system has no solutions, then the value of \(r\) must be \(-\frac{576}{37}\).

‍

The “\(x\) solutions” problem can be asked differently though. Rather than 2 equations, they may give a singular equation such as the example below:

‍

\[ a(-3x - 1) + x = 7x - 2 \]

‍

The equation above has no solutions, and \(a\) is a constant. What is the value of \(a\)?

‍

The strategy here remains very similar. You want to get both sides of the equation into slope-intercept form, then compare their slopes.

‍

\[ -3ax - a + x = 7x - 2 \]

\[ -3ax - a = 6x - 2 \]

‍

In this case, the slope of the right side of the equation would be equal to 6, while the left side of the equation has a slope of \(-3a\). You can now set these equal to each other (as that makes the lines parallel) in order to solve for \(a\):

‍

\[ -3a = 6 \]

\[ a = -2 \]

‍

[fs-toc-h3]Infinite Solutions

‍

Infinitely many solutions are not tested anywhere near as much on the DSAT compared to no solutions, but it is still important to know.

‍

If a system has infinitely many solutions, then the lines of both equations overlap on top of each other. In other words, when the equations are in slope-intercept form, both equations are equivalent (same slope and same y-intercept).

‍

‍

In the graph above, both lines lie on one another, meaning that there are an infinite number of \(x\) and \(y\) values that satisfy the equation above.

‍

For more on this concept, watch this, and this.

‍

[fs-toc-h2]Writing Linear Systems

‍

Writing linear systems of equations is no different than writing a singular linear equation (often in standard form).

‍

All the same concepts apply, you simply have to add on an extra equation.

‍

Take this one equation at a time.

‍

The first equation we’ll tackle deals with quantity.

‍

We’re told that a total of 350 tickets were sold. These tickets are made up of both bench (B) tickets, and Lawn (L) tickets. Meaning the sum of both \(B\) and \(L\) would equal the total amount of tickets, 350, giving us a first equation of:

‍

\[ B + L = 350 \]

‍

The next equation deals with price.

‍

Each bench ticket costs $75, meaning the total money made from bench tickets would be \(75 \cdot B\).

‍

Each lawn ticket costs $40, so the total money made from those would be \(40 \cdot L\).

‍

We’re also told that the total money made from both tickets comes out to $19,250. Making the final second equation:

‍

\[ 75B + 40L = 19,250 \]

‍

Scanning the answer choices, D is the one that matches up with both of our equations, making it the correct answer choice.

‍

In this case, not only do you have to write your own system, you must also solve it. All the steps are the same compared to the last question, other than the extra solving step added at the end.

‍

We’re not given any variables here to represent salads and drinks, so I'll assign the variable \(x\) to salads, and \(y\) to drinks.

‍

Why not use \(S\) for salads and \(D\) for drinks? You definitely can, but if you plan on solving the system by graphing on Desmos, \(x\) and \(y\) are the only variables it’ll identify. Using other letters such as \(S\) or \(D\) will result in an error.

‍

Firstly, we’re given that the total amount of \(x\) salads and \(y\) drinks sold in one day is 209. Put into an equation, this would equal:

‍

\[ x + y = 209 \]

‍

Next is our price equation. \(x\) salads cost $6.50 each, making the total money made from salads \(6.50x\). \(y\) drinks cost $2.00 each, making the total money made from drinks \(2.00y\).

‍

Since the sum of both of these is $836.50, the second equation is:

‍

\[ 6.50x + 2.00y = 836.50 \]

‍

Giving us a final system of:

‍

\[ x + y = 209 \]
\[ 6.50x + 2.00y = 836.50 \]

‍

All that's left to do is actually solve the system. The easiest thing to do here is to plug both equations into Desmos and find the intersection. The \(x\) value of the coordinate will be the number of salads in this case, and the \(y\) value will be the number of drinks.

‍

‍

The graph tells us that 93 salads were sold, making choice B the correct answer.

‍

The most important skill here is careful reading. There’s no real challenging computations here, it's simply matching up quantity with quantity and price with price.

‍

‍

[fs-toc-h2]Systems of Inequalities

‍

If you're interested in how to write a system of inequalities, check out the inequalities page. Here, we’ll be focused on solving a given system of inequalities using graphing.

‍

Just like a system of linear equations, solving systems of linear inequalities is much easier by graphing. Take, for example, the system:

‍

\[ 2x + 6y > 14 \]
\[ y < 4x + 9 \]

‍

‍

In the graph above, there are three shaded regions: a grey one, a red, and a shared, overlapping region in between the two.

‍

If you recall, the solutions to any single inequality lie within the shaded region. When graphing two inequalities, it's where the two shaded regions overlap.

‍

Meaning in the graph above:

‍

• \((-4, 2)\) is not a solution, as it lies outside of any shaded region.
‍

• \((0, 0)\) is not a solution, as it only lies in the purple region.
‍

• \((-2, 5)\) is not a solution, as it lies only in the green region.
‍

• \((4, 4)\) IS a solution, as it lies in the overlapping of both shaded regions.

‍

‍

‍

The first step here would be graphing the system.

‍

‍

We can see that the overlapping region is the one to the far left, in the negative plane.

‍

B is the one that matches up with this graph, making choice B the correct answer.

‍

The question wants us to find the greatest possible value for \(b\), which in this case represents the y-value. The easiest thing to do here is, again, to graph the system.

‍

‍

If you recall from the inequalities lesson, when your inequality is a “less than or equal to,” then any point ON the line is also a solution.

‍

This means that not only is the overlapping shaded region a solution, but the intersection of both of the lines is also a solution. The intersection has the greatest y-value out of all the solutions, making 750 the correct answer.

‍

While I do recommend solving this by graphing, if you can compute quickly enough, it's possible to solve algebraically.

‍

Just like in systems of linear equations, you can use substitution to create a singular equation. In this case, you can substitute \(y\) in the top equation with \(5x\) to get:

‍

\[ 5x \leq -15x + 3000 \]

\[ 20x \leq 3000 \]

\[ x \leq 150 \]

‍

You can then plug this \(x\) value back into one of the inequalities:

‍

\[ y \leq 5(150) \]

\[ y \leq 750 \]

‍

Giving you the same exact answer, 750.