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SAT · Math · Algebra
Systems of two linear equations in two variables is part of Algebra in the SAT Math section. The questions below are written in the style of the real test, each with the answer and a worked explanation.
Question types you will see here include word problem with two unknowns, parallel versus infinite solutions, solve for a combination and meaning of the intersection point.
Common traps: swapping which variable is which; using one equation twice; solving for x when y is asked.
Question 1
-2x - 3y = -2 -2x - 4y = -2 If (x, y) is the solution to the system of equations above, what is the value of x + y?
-1
0
1
2
Answer: C. 1
Solve by elimination or substitution. The solution is x = 1, y = 0. Check: -2(1) + -3(0) = -2 and -2(1) + -4(0) = -2. Therefore x + y = 1.
Question 2
3x - 3y = -12 x + y = -4 If (x, y) is the solution to the system of equations above, what is the value of y?
-4
0
1
2
Answer: B. 0
Solve by elimination or substitution. The solution is x = -4, y = 0. Check: 3(-4) + -3(0) = -12 and 1(-4) + 1(0) = -4. Therefore y = 0.
Question 3
A mountain trail is 10 kilometres long and has only two kinds of section. On steep sections the trail gains 120 metres of elevation per kilometre, and on gentle sections it gains 40 metres per kilometre. The total elevation gain over the whole trail is 760 metres. How many kilometres of the trail are steep?
Enter your own answer.
Answer: 4.5
Let s be the steep kilometres: 120s + 40(10 - s) = 760, so 400 + 80s = 760 and s = 4.5. Wrong routes: 5.5 is the length of the gentle sections, the other unknown; 76 is the average gain per kilometre, 760 ÷ 10, not a length.
Question 4
A roofing crew is tiling two identical houses. On one house, 320 m² has been tiled so far and the crew adds 20 m² per day. On the other house, 20 m² has been tiled so far and a second crew adds 30 m² per day. In how many days will the two houses have the same tiled area?
6
10
15
30
Answer: D. 30
Set 320 + 20d = 20 + 30d, so 300 = 10d and d = 30 (D). A divides 300 by the sum of the rates (50 gives 6) as if the crews were working toward each other; B divides 300 by 30 (the faster rate); C divides 300 by 20 (the slower rate).
Question 5
At an autumn fair, tickets to watch a trebuchet launch pumpkins cost $8 for adults and $5 for children. The fair sold 140 tickets for a total of $955. How many children's tickets were sold?
Enter your own answer.
Answer: 55
With a adult and c child tickets: a + c = 140 and 8a + 5c = 955. Since 5(a + c) = 700, the extra 3a = 255, so a = 85 and c = 140 − 85 = 55. Reporting 85 gives the adult tickets, which solves for the wrong variable.
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