A system of equations is two (or more) equations that share the same variables. You solve for the values of the variables that satisfy ALL equations simultaneously.
The SAT tests two main solution methods: substitution and elimination (addition/subtraction).
Best when one equation already has a variable isolated (or is easy to isolate).
Steps: 1. Solve one equation for one variable 2. Substitute that expression into the other equation 3. Solve for the remaining variable 4. Plug back in to find the other variable
Example: > y = 2x + 3 > 3x + y = 13
Step 1: y is already isolated (y = 2x + 3) Step 2: Substitute into the second equation: 3x + (2x + 3) = 13 Step 3: 5x + 3 = 13 → 5x = 10 → x = 2 Step 4: y = 2(2) + 3 = 7 Solution: (2, 7)
Best when neither equation has a variable conveniently isolated, or when the coefficients align for easy cancellation.
Steps: 1. Line up the equations 2. Multiply one or both equations so a variable's coefficients are equal (and opposite) 3. Add the equations — the variable cancels out 4. Solve for the remaining variable, then substitute back
Example: > 2x + 3y = 12 > 4x − 3y = 6
The y-coefficients are already opposites (+3 and −3). Add the equations: > (2x + 4x) + (3y − 3y) = 12 + 6 > 6x = 18 → x = 3 > Substitute: 2(3) + 3y = 12 → 6 + 3y = 12 → 3y = 6 → y = 2 > Solution: (3, 2)
The SAT also tests whether a system has 0, 1, or infinitely many solutions:
How to check:
Real-world example: A school store sells pencils for $0.50 each and notebooks for $2 each. One student buys 4 pencils and 2 notebooks for $6. Another buys 2 pencils and 3 notebooks for $7. How much does each item cost?
More interesting: "A coffee shop sells coffee for x dollars and tea for y dollars. A customer buys 3 coffees and 1 tea for $14. Another buys 1 coffee and 2 teas for $9. What are the prices?" > 3x + y = 14 > x + 2y = 9 (Solve by elimination or substitution → x = $3.80, y = $2.60)
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Quiz Questions:
Q1. Solve the system: y = 3x − 1 and 2x + y = 9
A) (1, 2) B) (2, 5) C) (3, 8) D) (4, 1)
Answer: B — Substitute y = 3x − 1 into 2x + y = 9: 2x + (3x − 1) = 9 → 5x − 1 = 9 → 5x = 10 → x = 2. Then y = 3(2) − 1 = 5. Solution: (2, 5).
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Q2. Solve by elimination: 3x + 4y = 10 and 3x − 2y = 4
A) x = 1, y = 2 B) x = 2, y = 1 C) x = 3, y = 0 D) x = 0, y = 2.5
Answer: B — Subtract the second equation from the first: (3x + 4y) − (3x − 2y) = 10 − 4 → 6y = 6 → y = 1. Substitute: 3x + 4(1) = 10 → 3x = 6 → x = 2. Solution: (2, 1).
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Q3. Which of the following systems has no solution?
A) y = 2x + 3 and y = 2x − 5 B) y = 3x + 1 and y = −3x + 1 C) y = x + 4 and y = −x + 4 D) y = 2x + 1 and y = 4x + 2
Answer: A — Both equations have slope 2 (parallel lines) but different y-intercepts (3 and −5). Parallel lines never intersect — no solution. Choice B has different slopes; lines will intersect. Same for C and D.
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Q4. At a bake sale, brownies cost $2 and cookies cost $1.50. A group spent $17 and bought a total of 10 items. How many brownies did they buy?
A) 3 B) 4 C) 5 D) 6
Answer: B — Let b = brownies, c = cookies. b + c = 10 and 2b + 1.5c = 17. From the first: c = 10 − b. Substitute: 2b + 1.5(10 − b) = 17 → 2b + 15 − 1.5b = 17 → 0.5b = 2 → b = 4.
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Q5. A system of two linear equations has infinitely many solutions. Which of the following must be true?
A) Both equations have slope 0 (horizontal lines) B) The two equations are essentially the same line — one is a multiple of the other C) The lines are perpendicular to each other D) One equation has a positive slope and the other has a negative slope
Answer: B — Infinitely many solutions means every point on one line is also on the other — the lines are identical. This happens when one equation is a scalar multiple of the other (e.g., y = 2x + 4 and 2y = 4x + 8 are the same line).