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Solving Systems of Two Linear Equations

Find the point that makes two linear equations true at the same time, by graphing, substitution, or elimination, and use systems to solve word problems.

Practice this skill 30 min Systems of two linear equations Leer en español
Key idea

What you'll learn

A system of linear equations is two equations that must both be true at the same time, for the same x and the same y. For example:

\begin{cases} y = x + 1 \\ y = -x + 5 \end{cases}

In this lesson you will learn three ways to solve a system — graphing, substitution, and elimination — how to recognize systems with no solution or infinitely many solutions, and how to set up a system from a word problem with two unknowns, like the number of adult and child tickets sold, or the price of a sandwich and the price of a drink.

Why it matters on the GED: Systems appear as "What is the solution to the system?", as "Which point is on both lines?", as graphs where you click the intersection point, and as word problems about tickets, mixtures, money, and comparing plans. Real life: "At how many miles do these two rental deals cost the same?" is a system, and so is "How much did I put in each account?"

Words to know

  • system of equations — two (or more) equations with the same variables, solved together.
  • solution of a system — an ordered pair (x, y) that makes both equations true ((2, 3) solves y = x + 1 and y = -x + 5).
  • ordered pair — a point written (x, y): the x-value first, the y-value second.
  • intersection point — where two lines cross on a graph; it is the solution of the system.
  • substitution — replacing a variable with an equal expression (y = 2x lets you write x + 2x instead of x + y).
  • elimination — adding or subtracting the equations so that one variable cancels out.
  • slope-intercept form — y = mx + b, where m is the slope and b is the y-intercept.
  • standard form — Ax + By = C (for example, 2x + 3y = 7).
  • parallel lines — lines with the same slope and different y-intercepts; they never meet (no solution).
  • same line (coincident lines) — two equations that describe the same line; they meet everywhere (infinitely many solutions).

All Math words in the glossary

What a solution is, and the graphing method

A solution of a system has to work in both equations. To check a point, substitute it into each equation.

Is (2, 3) a solution of y = x + 1 and y = -x + 5?

  • First equation: 3 = 2 + 1 ✓
  • Second equation: 3 = -2 + 5 ✓

Both are true, so (2, 3) is the solution. A point like (1, 2) works in the first equation (2 = 1 + 1) but not the second (2 \ne -1 + 5 = 4), so it is not a solution of the system.

Graphing method. Each linear equation is a line. Every point on a line makes that equation true, so the point that is on both lines — where they cross — makes both equations true.

Coordinate plane from -5 to 5 on both axes. Line y = x + 1 rises from (-5, -4) to (4, 5). Line y = -x + 5 falls from (0, 5) to (5, 0). The lines cross at the point (2, 3), which is marked with a dot and labeled as the solution.-5-5-4-4-3-3-2-2-1-11122334455xyy = x + 1y = −x + 5(2, 3)
  1. Write each equation in y = mx + b form.
  2. Graph each line: plot the y-intercept b, then use the slope m (rise over run) to find more points.
  3. Read the intersection point (x, y).
  4. Check it in both equations.

Graphing is great for seeing the answer, and the GED often shows the graph for you. But if the intersection is not on a grid point (like (2.4, 1.7)), you need algebra.

How many solutions? Compare the slopes and intercepts in y = mx + b form:

The two lines…SlopesInterceptsNumber of solutions
cross oncedifferentanyexactly one
are parallelsamedifferentnone
are the same linesamesameinfinitely many

Example: y = 3x + 1 and y = 3x - 2 have the same slope (3) but different intercepts → parallel → no solution.

Three small graphs. Left: two lines with different slopes cross at one point, so there is exactly one solution. Middle: y = 3x + 1 and y = 3x − 2 have the same slope and different intercepts, so they are parallel and there is no solution. Right: two equations for the same line lie on top of each other, so there are infinitely many solutions.One solutiondifferent slopesNo solutionsame slope, parallelInfinitely manysame liney = 3x + 1 and y = 3x − 2

The substitution method

Use substitution when one equation is already solved for a variable (like y = 3x - 4 or x = 2y), or can easily be.

  1. Solve one equation for one variable (if it isn't already).
  2. Substitute that expression into the other equation. Use parentheses!
  3. Solve the new one-variable equation.
  4. Back-substitute to find the other variable.
  5. Check the pair in both original equations.

Example: \begin{cases} y = 3x - 4 \\ 2x + y = 11 \end{cases}

  • Replace y in the second equation with (3x - 4): 2x + (3x - 4) = 11.
  • Combine: 5x - 4 = 11, so 5x = 15 and x = 3.
  • Back-substitute: y = 3(3) - 4 = 5.
  • Solution (3, 5). Check: 2(3) + 5 = 11 ✓

Example with "equal" expressions: if y = x + 2 and y = 3x - 4, both right sides equal y, so they equal each other: x + 2 = 3x - 4. Then 6 = 2x, x = 3, and y = 3 + 2 = 5.

Careful with negatives: substituting y = 2x - 5 into 4x - y = 9 gives 4x - (2x - 5) = 9, which is 4x - 2x + 5 = 9. Without the parentheses you would get the wrong sign on the 5.

The elimination method

Use elimination when both equations are in standard form Ax + By = C.

  1. Line up the x terms, y terms, and constants.
  2. If needed, multiply one or both equations (every term!) so that one variable has opposite coefficients, like +6y and -6y.
  3. Add the equations. One variable cancels.
  4. Solve for the remaining variable, then back-substitute.

Example (ready to add): \begin{cases} 3x + 2y = 16 \\ 5x - 2y = 0 \end{cases}

The y terms are +2y and -2y. Add: 8x = 16, so x = 2. Back-substitute: 3(2) + 2y = 16, so 2y = 10 and y = 5. Solution (2, 5).

Coordinate plane with x from 0 to 6 and y from 0 to 9. Line 3x + 2y = 16, which is y = 8 − 1.5x, falls through (0, 8) and (4, 2). Line 5x − 2y = 0, which is y = 2.5x, rises through the origin. The lines cross at (2, 5), the solution found by adding the equations.012345601234567893x + 2y = 165x − 2y = 0(2, 5)

Example (multiply first): \begin{cases} 4x + 3y = 10 \\ 5x - 2y = 1 \end{cases}

The LCM of 3 and 2 is 6. Multiply the first equation by 2 and the second by 3:

\begin{cases} 8x + 6y = 20 \\ 15x - 6y = 3 \end{cases}

Add: 23x = 23, so x = 1. Then 4(1) + 3y = 10, so y = 2. Solution (1, 2).

Subtracting instead of adding. If a variable has the same coefficient in both equations (like 2x + y = 11 and 2x - 3y = -1), subtract one equation from the other. Be careful to subtract every term. Many people prefer to multiply one equation by -1 and then add.

Special cases. If both variables cancel:

  • a false statement (like 0 = 7) → no solution (parallel lines);
  • a true statement (like 0 = 0) → infinitely many solutions (the same line).

Word problems with two unknowns. Name both unknowns, write one equation for each fact in the story, then solve. Typical pairs of facts: a count (a + c = 200 tickets) and a value (8a + 5c = 1390 dollars); or a sum and a difference; or "with the current" and "against the current".

Formula

Forms to know

There is no special "systems formula". You use the two forms of a line:

y = mx + b \qquad\qquad Ax + By = C
  • In y = mx + b: m = slope, b = y-intercept. This form is on the GED formula sheet (along with the slope formula m = \dfrac{y_2 - y_1}{x_2 - x_1}).
  • Ax + By = C is standard form; A, B, C are numbers. You need to recognize it — it is not listed on the sheet.

For word problems, these sheet formulas often supply the two equations: d = rt, I = prt, and total cost = (number of units) \times (price per unit).

Memorize: same slope + different intercept → no solution; same slope + same intercept → infinitely many solutions; different slopes → exactly one solution.

Worked example

Example 1 (easy): read and verify a graph

The graph in the first section of this lesson shows y = x + 1 and y = -x + 5. What is the solution of the system?

  1. Read the crossing point from the graph: (2, 3).
  2. Confirm with algebra: both equal y, so x + 1 = -x + 5.
  3. Add x to both sides: 2x + 1 = 5. Subtract 1: 2x = 4, so x = 2.
  4. y = 2 + 1 = 3.
Two tables of values for x = 0 to 4. For y = x + 1 the y-values are 1, 2, 3, 4, 5. For y = −x + 5 the y-values are 5, 4, 3, 2, 1. Only the row x = 2 has the same y-value, 3, in both tables, so (2, 3) is the solution.xy = x + 1y = −x + 5015124233342451same y!At x = 2 both equations give y = 3.x + 1 = −x + 5 → 2x = 4 → x = 2, y = 3Solution: (2, 3) — the point where the lines cross

Answer: (2, 3)

Check: 3 = 2 + 1 ✓ and 3 = -2 + 5 ✓

Worked example

Example 2: substitution

Solve \begin{cases} y = 3x - 4 \\ 2x + y = 11 \end{cases}

  1. The first equation gives y. Substitute (3x - 4) for y in the second: 2x + (3x - 4) = 11.
  2. Combine like terms: 5x - 4 = 11.
  3. Add 4: 5x = 15. Divide by 5: x = 3.
  4. Back-substitute: y = 3(3) - 4 = 9 - 4 = 5.
Coordinate plane with x from 0 to 6 and y from 0 to 12. Line y = 3x − 4 rises steeply through (2, 2) and (4, 8). Line 2x + y = 11, which is y = −2x + 11, falls through (0, 11) and (5, 1). The lines cross at (3, 5), the solution.0123456024681012y = 3x − 42x + y = 11(3, 5)

Answer: (3, 5)

Check: 5 = 3(3) - 4 ✓ and 2(3) + 5 = 11 ✓

Worked example

Example 3: elimination after multiplying

Solve \begin{cases} 4x + 3y = 10 \\ 5x - 2y = 1 \end{cases}

  1. Make the y coefficients opposites (+6y and -6y). Multiply the first equation by 2: 8x + 6y = 20. Multiply the second by 3: 15x - 6y = 3.
  2. Add the equations: 23x = 23, so x = 1.
  3. Back-substitute into 4x + 3y = 10: 4 + 3y = 10, so 3y = 6 and y = 2.
Coordinate plane with x from 0 to 4 and y from 0 to 6. Line 4x + 3y = 10 falls from (0, 3⅓) to (2.5, 0). Line 5x − 2y = 1 rises from (0.2, 0) through (1, 2), reaching y = 6 at x = 2.6. The lines cross at (1, 2), the solution.0123401234564x + 3y = 105x − 2y = 1(1, 2)

Answer: (1, 2)

Check: 4(1) + 3(2) = 10 ✓ and 5(1) - 2(2) = 1 ✓

Worked example

Example 4 (real world): sandwiches and drinks

At a café, 3 sandwiches and 2 drinks cost $27.50. 2 sandwiches and 4 drinks cost $25.00. What is the price of one sandwich?

  1. Let s = price of a sandwich and d = price of a drink.
  2. Two facts, two equations: 3s + 2d = 27.50 and 2s + 4d = 25.00.
  3. Multiply the first equation by 2: 6s + 4d = 55.00.
  4. Subtract the second equation: (6s + 4d) - (2s + 4d) = 55.00 - 25.00, so 4s = 30 and s = 7.50.
  5. Back-substitute: 3(7.50) + 2d = 27.50, so 22.50 + 2d = 27.50, 2d = 5, d = 2.50.
Graph of the café system. Horizontal axis: sandwich price s in dollars from 0 to 14. Vertical axis: drink price d in dollars from 0 to 14. Line 3s + 2d = 27.50 runs from (0, 13.75) down to about (9.17, 0). Line 2s + 4d = 25.00 runs from (0, 6.25) down to (12.5, 0). They cross at (7.5, 2.5): a sandwich costs 7.50 dollars and a drink 2.50 dollars.0246810121402468101214Sandwich price s ($)Drink price d ($)3s + 2d = 27.502s + 4d = 25.00(7.50, 2.50)

Answer: a sandwich costs $7.50 (and a drink $2.50)

Check: 2(7.50) + 4(2.50) = 15 + 10 = \$25.00 ✓

Worked example

Example 5 (GED-hard, work backwards): two accounts

Mr. Lee invested a total of $10,000 in two accounts. One pays 3% simple interest per year and the other pays 5%. After one year he earned $380 in interest. How much did he put in the 3% account?

  1. Let x = dollars at 3% and y = dollars at 5%.
  2. Amount fact: x + y = 10{,}000, so y = 10{,}000 - x.
  3. Interest fact (using I = prt with t = 1): 0.03x + 0.05y = 380.
  4. Substitute: 0.03x + 0.05(10{,}000 - x) = 380, so 0.03x + 500 - 0.05x = 380.
  5. Combine: 500 - 0.02x = 380. Subtract 500: -0.02x = -120. Divide: x = 6000.
  6. Then y = 10{,}000 - 6000 = 4000.
Graph of the two-accounts system. Horizontal axis: dollars x in the 3% account, 0 to 14,000. Vertical axis: dollars y in the 5% account, 0 to 12,000. Line x + y = 10,000 runs from (0, 10,000) to (10,000, 0). Line 0.03x + 0.05y = 380 runs from (0, 7,600) to about (12,667, 0). They cross at (6,000, 4,000).02k4k6k8k10k12k14k02k4k6k8k10k12kx = dollars at 3%y = dollars at 5%x + y = 10,000 (amount)0.03x + 0.05y = 380 (interest)(6,000, 4,000)

Answer: $6,000 in the 3% account (and $4,000 in the 5% account)

Check: 0.03(6000) + 0.05(4000) = 180 + 200 = \$380 ✓

Calculator tip

Calculator tip (TI-30XS MultiView)

The TI-30XS MultiView does not solve systems for you on the GED — you set up and eliminate on paper. Use the calculator to:

  • Check an answer choice quickly. For (5, -1) in 3x - 4y = 19, type 3 × 5 − 4 × (−) 1 enter → 19. Use the (−) key for the negative 1.
  • Handle decimal word problems. For 0.03(6000) + 0.05(4000) type 0.03 × 6000 + 0.05 × 4000 enter → 380.
  • Divide messy numbers after elimination, such as 23x = 23 or -0.02x = -120 (type (−) 120 ÷ (−) 0.02 → 6000).

Fastest strategy on multiple choice: substitute each answer choice into both equations with the calculator. Only the correct pair works in both — and wrong choices are often built to work in just one.

Common mistake

Common mistakes

  • Checking only one equation. (4, 2) works in 3x + 2y = 16 but not in 5x - 2y = 0. A solution must work in both.
  • Swapping x and y. If x = 2 and y = 5, the answer is (2, 5), not (5, 2). The x-value always comes first.
  • Multiplying only one side or one term. To multiply 4x + 3y = 10 by 2, multiply every term: 8x + 6y = 20, not 8x + 6y = 10.
  • Subtracting only some terms. When subtracting equations, subtract the constant too, and watch double negatives: (5x - 2y) - (3x - 2y) = 2x, because -2y - (-2y) = 0.
  • Dropping parentheses in substitution. 4x - (2x - 5) is 2x + 5, not 2x - 5.
  • Stopping halfway. After finding x, you still need y — and you need to answer the quantity the word problem asks for.
Test strategy

GED test strategy

  • Plug in the choices. For "What is the solution?" questions, test each ordered pair in both equations. This is often the fastest method and is almost error-proof.
  • Pick the easiest method. Already solved for y? Use substitution. Both in standard form with matching or opposite coefficients? Use elimination. Graph given? Read the intersection and verify.
  • On hot-spot graph questions, click exactly where the lines cross — not where a line meets an axis.
  • No-solution / infinitely-many questions: rewrite both in y = mx + b form and compare slopes and intercepts.
  • Word problems: write one equation per fact. Look for a "count" fact and a "value" fact. Then make sure you answer the variable the question asks for.
  • Estimate: in a mixture of $9 and $14 coffee making a $12 blend, the blend is closer to $14, so there must be more of the $14 coffee.

Check yourself

Systems of two linear equations

Which ordered pair is the solution to the system of equations?

\begin{cases} y = x + 2 \\ y = 3x - 4 \end{cases}
Answer choices
Hint

Both equations equal y, so set the right sides equal. Or test each choice in both equations.

Check yourself

Systems of two linear equationsHot spot

The graph shows the lines y = -x + 3 and y = 2x - 3.

Click the point that represents the solution of the system.

Click on the figure to mark your answer (or focus it and use the arrow keys).

Coordinate plane from -5 to 5 on both axes showing two lines. Line 1 is y = -x + 3: it crosses the y-axis at (0, 3) and goes down to the right through (3, 0). Line 2 is y = 2x - 3: it crosses the y-axis at (0, -3) and rises steeply through (1, -1), (3, 3) and (4, 5). The two lines cross at one point.-5-5-4-4-3-3-2-2-1-11122334455xyy = −x + 3y = 2x − 3
Hint

The solution is the point that lies on both lines at once.

Summary

  • The solution of a system is the (x, y) pair that makes both equations true — the point where the lines cross.
  • Substitution: solve one equation for a variable, substitute (with parentheses) into the other.
  • Elimination: multiply so one variable has opposite coefficients, add, solve, back-substitute.
  • Same slope, different intercepts → no solution; same line → infinitely many; different slopes → one solution.
  • In word problems, name both unknowns and write one equation for each fact; check your answer in the story.
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