Solving Linear Word Problems
Once the equation is set up, the arithmetic is usually easy — the points are lost on units and on answering the wrong question.
Reading the concept isn't enough. The score comes from practicing the real Bluebook-style format and finding the exact mistakes you keep making.
What the SAT tests here
Keep units consistent (convert before you compute, not after), and re-read what the question asks for. The SAT loves to make you find $x$ when it actually wants $2x + 1$.
How to solve one, step by step
Example: a car travels at $60$ miles per hour. How far in $2.5$ hours?
- Distance = rate $\times$ time: $60 \times 2.5$.
- That's $150$ miles.
The mistakes that cost points
- Ignoring unit conversions. Minutes vs. hours, cents vs. dollars — convert before computing.
- Answering the wrong quantity. Solve for exactly what the question asks, not an intermediate value.
Practice questions
Try these the way you would on test day, then open the solution to check your method.
Easy
The graphs of $y = 2x + k$ and $y = -x + 6$ intersect at the point where $x = 2$. What is the value of $k$?
- A$k = -2$
- B$k = 4$
- C$k = 0$
- D$k = 2$
Show solution
Answer: C, $k = 0$. Since the lines intersect at $x = 2$, both equations share the same $y$-value there. First find $y$ using the second equation: $y = -(2) + 6 = 4$. Now substitute $x = 2$ and $y = 4$ into the first equation: $4 = 2(2) + k$, so $4 = 4 + k$, giving $k = 0$.
Medium
The graph of $y = k$ is a horizontal line, and the graph of $y = -2x + 8$ passes through the same point when $x = 3$. At what value of $k$ do the two graphs intersect?
- A$k = 8$
- B$k = 3$
- C$k = -2$
- D$k = 2$
Show solution
Answer: D, $k = 2$. The line $y = k$ is horizontal. To find where it intersects $y = -2x + 8$ at $x = 3$, substitute $x = 3$ into the second equation: $y = -2(3) + 8 = -6 + 8 = 2$. Since $y = k$ at every point on the horizontal line, the intersection requires $k = 2$.
Hard
In the $xy$-plane, the lines $y = \frac{1}{2}x + k$ and $y = -3x + 11$ intersect at a point where $x = 4$. What is the value of $k$?
- A$k = -3$
- B$k = 11$
- C$k = 2$
- D$k = -1$
Show solution
Answer: A, $k = -3$. At $x = 4$, substitute into $y = -3x + 11$: $y = -3(4) + 11 = -12 + 11 = -1$. The intersection point is $(4, -1)$. Substitute $(4, -1)$ into $y = \frac{1}{2}x + k$: $-1 = \frac{1}{2}(4) + k = 2 + k \Rightarrow k = -3$.
Find your exact gaps
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