SAT Math study guide · Chapter 6
Systems of Two Linear Equations
A system combines two constraints. The task is to determine what those constraints fix, whether that is a point, a variable, or a requested combination.
You saw in the craft chapters that a single relationship can fix a combination of variables without fixing the variables themselves. A system is where that idea earns the most, because now you hold two relationships at once, and two relationships can do only a few things. They can pin the variables to a single shared point; they can agree so completely that the second adds nothing the first did not already say; or they can contradict each other so that nothing satisfies both. Reading which of these is happening, and extracting exactly what was asked with the least fragile work, is what this section trains.
▸ Solve systems by elimination or substitution
Elimination and substitution are two standard routes. Choose the one that extracts the requested information with the least unnecessary work. Substitution shines when one variable is already alone, or nearly so: you replace it and collapse two equations into one.
WORKED By substitution
y = 2x - 3 and 3x + y = 12. The first equation already gives y alone, so substitute it into the second.
3x + (2x - 3) = 12
5x - 3 = 12 → 5x = 15 → x = 3
Then y = 2(3) - 3 = 3. The solution is the single pair (3, 3).
Elimination shines when the coefficients already line up, or can be scaled to line up: you add or subtract the equations to cancel one variable.
WORKED By elimination
4x + 3y = 10 and 4x - y = 2. The x-terms are identical, so subtracting removes x outright.
(4x + 3y) - (4x - y) = 10 - 2
4y = 8 → y = 2
Back-substitute: 4x + 3(2) = 10, so 4x = 4 and x = 1. The solution is (1, 2).
Choose the route after you have seen what the question wants. If it asks for a single variable, or for a combination, you may not need the full solution at all, which is the next idea.
Hand / Desmos / Hybrid: a system with awkward-looking coefficients is a strong Desmos candidate: graph both equations and click the intersection, and both x and y are read off directly with no elimination arithmetic at all. Clean, small-integer coefficients are usually just as fast by hand; the messier the numbers, the more the graph route pays off.
▸ Use linear combinations to find a requested expression
Often the requested quantity is a combination of the variables, and a combination can be assembled straight from the two equations without ever finding x and y separately. The cleanest move is to look at the target first and ask whether adding, subtracting, or scaling the equations produces it directly.
WORKED Assemble the target by adding
2x + y = 9 and x + 3y = 12. Find 3x + 4y.
Add the two equations and the requested combination simply appears:
(2x + y) + (x + 3y) = 9 + 12 → 3x + 4y = 21
No coordinate pair was needed. This avoids unnecessary algebra and reduces opportunities for sign errors.
Sometimes the target needs a little scaling before it falls out. To find x alone from x - y = 1 and 2x + 3y = 12, scale the first equation by 3 so the y-terms cancel on adding:
3(x - y) = 3 → 3x - 3y = 3, then (3x - 3y) + (2x + 3y) = 3 + 12 → 5x = 15, x = 3.
But do not force a combination that does not fit. If the target does not line up cleanly with the equations, scaling them into knots is more fragile than simply solving the system the ordinary way. The judgment is one quick test: can the target be built directly? If yes, build it; if no, solve enough and no more.
▸ Classify systems as one solution, no solution, or infinitely many
When a system is solved fully, the result is one of three outcomes, and each is the same statement told twice: once in algebra, once in geometry. Two independent equations meet at a single shared point: one solution.
WORKED One solution
2x + y = 5 and x - y = 1. Add to remove y:
3x = 6 → x = 2, then y = x - 1 = 1.
The lines cross exactly once, at (2, 1).
FIGURE 1 Independent equations cross at a single point, exactly one pair satisfies both.
Two equations describing parallel lines never meet: no solution. In algebra this shows up as the same left-side combination being assigned two different values, the larger cousin of the contradiction you met when a single equation's variable cancelled.
WORKED No solution
x + 2y = 4 and 2x + 4y = 9. Double the first equation to compare:
2x + 4y = 8, but the second says 2x + 4y = 9.
The same combination cannot equal both 8 and 9, so nothing satisfies both. The lines are parallel.
FIGURE 2 Parallel lines never intersect: one combination is asked to equal two different numbers.
And two equations that are secretly the same line agree everywhere: infinitely many solutions. Here the second equation is just a scaled copy of the first, adding no new restriction.
WORKED Infinitely many solutions
x + 2y = 4 and 2x + 4y = 8. The second is exactly twice the first.
Eliminating gives 0 = 0: a statement that is always true, which is the system reporting that the two equations carry one fact between them.
FIGURE 3 One line drawn twice: every point on it satisfies both equations.
This gives a clean reading rule. Eliminate a variable and look at what is left: a value for the remaining variable means one solution; a true statement with no variable, like 0 = 0, means infinitely many; a false statement, like 0 = 5, means none. You never have to graph anything to tell them apart.
The three system outcomes
Lines cross once → one solution → elimination leaves a value.
Lines are parallel → no solution → elimination leaves a false statement.
Lines coincide → infinitely many → elimination leaves a true statement.
▸ Find a constant that creates one, no, or infinitely many solutions
When a letter sits among the coefficients, it is usually not another variable to solve for; it selects which of the three outcomes the system has. The reading is structural: if the parameter makes the variable terms proportional, the constants then decide between no solution and infinitely many; if it leaves them non-proportional, the system has exactly one solution.
WORKED Choosing a parameter for no solution
For what value of a does ax + 4y = 7 and 6x + 8y = 5 have no solution?
No solution needs the left sides proportional but the constants out of step. The y-coefficients 4 and 8 are in ratio 1 : 2, so the x-coefficients must match that ratio:
a / 6 = 1 / 2 → a = 3.
Check the constants at a = 3: the first equation is 3x + 4y = 7, while half of the second is 3x + 4y = 2.5. The same combination is assigned two values, so there is no solution; exactly what was wanted.
Notice the half-step that is easy to skip: matching the coefficient ratio is necessary but not the end. Had the constants also matched that ratio, the very same a = 3 would have produced infinitely many solutions instead. The parameter sets the slopes equal; the constants decide whether the two lines are parallel or identical.
▸ Create and interpret systems from context
A word problem usually hands you two facts about two quantities, and asks for one quantity, a total, or a difference. Translate each fact into an equation, then choose the lightest route to the thing requested.
WORKED A concession stand
Popcorn costs 4 dollars and a drink costs 3 dollars. One afternoon the stand sold 50 items for 172 dollars. How many drinks were sold?
Let p be popcorns and d be drinks. The two facts are a count and a total:
p + d = 50 4p + 3d = 172
Substitute p = 50 - d into the total: 4(50 - d) + 3d = 172, so 200 - d = 172 and d = 28 (with p = 22).
Both counts are whole and nonnegative, so the answer fits the situation: 28 drinks.
WORKED Reading both coordinates of an intersection
Shipping plan A costs 20 dollars plus 3 dollars per pound; plan B costs 10 dollars plus 5 dollars per pound. At what weight do the two plans cost the same, and what is that shared cost?
Set the two cost expressions equal and solve for the weight:
20 + 3x = 10 + 5x → x = 5
The intersection is a point, not just an x-value, and the question asked about both of its coordinates. The weight is 5 pounds; the shared cost is 20 + 3(5) = 35 dollars. Reporting only x = 5 is incomplete because the question asks for both the weight and the shared cost, represented by the point (5, 35).
That last check is not a formality. A real-world system carries restrictions the algebra does not know about: counts must be whole and nonnegative, lengths must be positive, a probability must sit between zero and one. If the algebra ever returns a value that the situation cannot hold, first rule out a setup or arithmetic error before drawing a conclusion; only after ruling out an error does the situation itself have no valid answer, and recognizing that is part of what the question is testing.
▸ Use answer choices to check the target
System answer choices are usually built from the levels of information the question is sorting through. If the target is x + y, the choices often include x, y, x - y, and the ordered pair: each a true by-product of correct work, only one of them the requested object. If the target is a parameter that gives no solution, expect the value that instead gives infinitely many to be sitting right beside it. The familiar-looking answer is frequently an honest intermediate value, not the thing that was asked. Use the answer choices as a final target check, but report only the quantity named in the stem.
▸ Practice: systems of linear equations
Hard transfer: questions 5 and 9 ask for the parameter itself rather than a solution, the harder direction of the solution-count skill above.
Question 1
- A. 2
- B. 5
- C. 3
- D. 4
Question 2
- A. 2
- B. 3
- C. 4
- D. 6
3. If 2x + 5y = 11 and 3x - 5y = 4, what is the value of 5x? (grid-in)
Question 4
- A. 12
- B. 0
- C. 6
- D. any value except 12
5. For what value of a does ax + 4y = 7 and 6x + 8y = 5 have no solution? (grid-in)
Question 6
- A. exactly one
- B. none
- C. infinitely many
- D. exactly two
Question 7
- A. exactly one
- B. none
- C. infinitely many
- D. exactly two
8. Line A is y = 3x + 2. Line B passes through (1, 4) and (4, 13). How many solutions does the system of these two lines have? (grid-in)
9. The system ax + 6y = 10 and 4x + 12y = 20 has infinitely many solutions. What is the value of a? (grid-in)
10. Adult tickets cost 10 dollars and child tickets cost 6 dollars. A group bought 40 tickets for 304 dollars. How many adult tickets were bought? (grid-in)
Question 11
- A. exactly one
- B. none
- C. infinitely many
- D. exactly two
12. If 3x + y = 7 and x + y = 3, what is the value of x? (grid-in)
13. Plan A costs C = 20 + 8x and Plan B costs C = 5 + 12x. For what value of x are the two costs equal? (grid-in)
Question 14
- A. 4
- B. 9
- C. 13
- D. (4, 9)
Question 15
- A. 0
- B. 9
- C. 18
- D. cannot be determined
16. If x + 2y = 9 and 3x + 6y = 27, what is the value of x + 2y? (grid-in)
17. For what value of k does 2x + 3y = 12 and 4x + ky = 25 have no solution? (grid-in)
▸ Solutions: systems of linear equations
Show the answers and explanations
1: A. Substitute the first equation into the second: 2x + (3x - 1) = 9, so 5x = 10 and x = 2. Choice 5 is the value of y, the variable not asked for.
2: B. Add the equations to remove y: 8x = 16, so x = 2; then 5(2) + 2y = 16 gives y = 3. Adding was available because the y-terms were already opposites.
3: 15. Add the two equations and the y-terms cancel: (2x + 5y) + (3x - 5y) = 11 + 4, so 5x = 15. The question asked for 5x, which is delivered without finding x.
Show the answers and explanations
4: D. The left sides are identical, so the constants decide. If k = 12 the equations are the same line (infinitely many solutions); for the system to have no solution the constants must differ, so k can be any value other than 12. Choice 12 produces the opposite outcome.
5: 3. No solution needs proportional coefficients with mismatched constants. The y-coefficients 4 and 8 are in ratio 1 : 2, so a/6 = 1/2 gives a = 3. (At a = 3 the constants 7 and 2.5 do not match the ratio, confirming no solution.)
Show the answers and explanations
6: C. The second equation is exactly twice the first, so they describe the same line and share infinitely many solutions. Eliminating leaves 0 = 0.
7: B. Doubling the first gives 2x - 2y = 6, but the second says 2x - 2y = 1. The same combination cannot be both, so there is no solution.
8: none. Line A has slope 3 and intercept 2. Line B has slope (13 - 4)/(4 - 1) = 3 and, using the point (1, 4), intercept 1. Equal slopes with different intercepts mean the lines are parallel, so the system has no solution.
9: 2. Infinitely many solutions means the second equation is a scaled copy of the first. Since 4x + 12y = 20 is twice 2x + 6y = 10, the value of a must match: a = 2.
10: 16. Let a + c = 40 and 10a + 6c = 304. Substituting c = 40 - a gives 10a + 6(40 - a) = 304, so 4a + 240 = 304 and a = 16. Both counts are whole and nonnegative, so the answer fits.
Show the answers and explanations
11: C. The second equation is twice the first, so the two are one line: infinitely many solutions.
12: 2. Subtract the second equation from the first: (3x + y) - (x + y) = 7 - 3, so 2x = 4 and x = 2. Subtraction removed y in a single step.
13: 3.75. The costs are equal where 20 + 8x = 5 + 12x, so 15 = 4x and x = 3.75. This is the crossing point left open in the previous section, now found by setting the two models equal, a system of two equations.
Show the answers and explanations
14: C. The requested object is the sum: x + y = 4 + 9 = 13. Choices 4 and 9 are the individual variables, and (4, 9) is the full solution: all true by-products, none of them the sum that was asked for.
15: D. The second equation is exactly twice the first, so it repeats the same fact rather than adding a new one; the system carries only one genuine restriction on x and y. Infinitely many pairs satisfy it, and different pairs give different values of x - y, so it cannot be determined.
16: 9. The second equation is exactly three times the first, so the pair (x, y) is not unique. But the requested combination, x + 2y, is fixed by the first equation alone: it equals 9 no matter which solution pair is chosen.
17: 6. No solution needs proportional coefficients with mismatched constants. The x-coefficients 2 and 4 are in ratio 1 : 2, so k must match: 3/k = 1/2 gives k = 6. (At k = 6, doubling the first equation gives 4x + 6y = 24, which conflicts with the constant 25, confirming no solution.)
▸ Systems of linear equations: chapter summary
Two linear equations can intersect once, coincide, or be parallel. Elimination identifies the outcome quickly, and a requested linear combination can often be obtained without solving for both variables. Parameters control whether coefficients and constants become proportional. In context problems, check that the resulting values satisfy the stated restrictions.