How to solve SAT systems of linear equations

A system of two linear equations asks for the pair $(x, y)$ that makes both equations true at once. Graphically, that pair is where the two lines cross. On the Digital SAT you solve systems directly, set them up from word problems, and decide when a system has no solution or infinitely many. Algebra is about 35% of the Math section, or 13 to 15 of the 44 questions, and systems usually account for roughly 2 to 4 of them.

What the questions look like

Solve the system. You get two equations and must find $x$, $y$, or a combination such as $x + y$ or $x - y$. The combination versions often have a shortcut.

Set up from a story. Two unknown quantities, like numbers of two kinds of tickets or amounts of two solutions, are tied together by a total count and a total value. You write both equations and solve for the requested quantity.

Count the solutions. One equation contains a constant such as $k$, and you find the value that makes the system have no solution or infinitely many solutions.

Step-by-step method

1. Line up both equations in the same form, ideally $ax + by = c$.
2. Pick a method. If one variable is already isolated, like $y = 3x - 2$, substitute it into the other equation. Otherwise use elimination. Multiply one or both equations so a variable has opposite coefficients, then add.
3. Look for a shortcut when the question asks for a combination. If $2x + 3y = 17$ and $4x + 5y = 31$, subtracting the first equation from the second gives $2x + 2y = 14$, so $x + y = 7$ without solving for either variable.
4. Back-substitute to find the second variable if you need it.
5. For solution-count questions, write both equations as $ax + by = c$ and compare the ratios of matching coefficients.
   • If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, there is exactly one solution.
   • If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel and there is no solution.
   • If all three ratios are equal, the equations describe the same line and there are infinitely many solutions.
6. Check your answer in both original equations.

Worked example

Example 1. A cafe sold 120 drinks in one morning. Each coffee cost $3.50 and each smoothie cost $5.75. The cafe collected $501 from these drinks. How many smoothies did it sell?

Let $c$ be the number of coffees and $s$ the number of smoothies. One equation counts drinks and the other counts dollars:

$$c + s = 120$$
$$3.5c + 5.75s = 501$$

From the first equation, $c = 120 - s$. Substitute:

$$3.5(120 - s) + 5.75s = 501$$
$$420 - 3.5s + 5.75s = 501$$
$$2.25s = 81$$
$$s = 36$$

The cafe sold 36 smoothies, which means 84 coffees. Check the money: $84(3.50) + 36(5.75) = 294 + 207 = 501$. It matches.

Example 2. In the system below, $k$ is a constant.

$$3x + ky = 8$$
$$6x + 10y = 20$$

For what value of $k$ does the system have no solution?

For no solution, the $x$ and $y$ coefficients must be in the same ratio while the constants are not. The $x$ ratio is $\frac{3}{6} = \frac{1}{2}$, so the $y$ ratio must also be $\frac{1}{2}$:

$$\frac{k}{10} = \frac{1}{2} \quad \Rightarrow \quad k = 5$$

Now check the constants: $\frac{8}{20} = \frac{2}{5}$, which is not $\frac{1}{2}$. The lines are parallel and distinct, so $k = 5$ gives no solution. If the first constant had been 10 instead of 8, $k = 5$ would make the equations identical and give infinitely many solutions.

Common traps

• Solving for the wrong variable. In Example 1, the number of coffees, 84, would be a likely wrong choice. Before you answer, reread which quantity the question names.
• Subtracting only part of an equation. In elimination, subtract every term, including the constant. A sign error on one term is the most common way to get a wrong pair.
• Confusing no solution with infinitely many. Parallel lines with different intercepts never meet. Identical lines meet everywhere. Check the constants after you match the coefficients.
• Mixing up the count equation and the value equation. In money and mixture problems, one equation adds the quantities and the other adds price times quantity. Don't put prices into the count equation.
• Using whole-number percents in mixtures. A 30% solution contributes $0.30x$, not $30x$, to the amount of pure substance.

Desmos tip

Graph both equations as written and click the point where the lines cross. Desmos labels the exact coordinates, which handles decimal answers like $(2.4, -1.75)$ that are slow by hand. For a question with $k$, type the equation with $k$ in it, add the slider, and drag until the lines become parallel or overlap. Then confirm the value algebraically, since slider steps can skip the exact number.