Lesson 7 of 35 · Quantitative Reasoning: algebra

Linear equations and systems

Solve one- and two-variable linear equations by hand, combine equations to reach the asked quantity directly, and spot systems with no unique solution.

About 20 minutes

In this lesson

  • Solve a linear equation in one variable without slips in sign or order.
  • Solve a two-variable system by elimination or substitution, choosing the faster route.
  • Reach a combination such as x+yx + y or 3x+4y3x + 4y straight from the equations, without solving for each variable.
  • Tell when two equations give one solution, none or infinitely many.

The idea

Why linear algebra is worth fast hands

GMAT Quantitative Reasoning is 21 problem-solving questions in 45 minutes, and you cannot use a calculator in that section. Linear equations appear on their own and inside nearly every word problem, so the gain is not in knowing how to solve them (you do) but in solving them in a few lines, with no sign slips, and in noticing when the question lets you skip most of the work.

One variable: undo in reverse order

An equation stays true if you do the same thing to both sides. To isolate xx, undo what was done to it in reverse order: first remove what was added, then what was multiplied.

3x−45=7  ⇒  3x−4=35  ⇒  3x=39  ⇒  x=13\frac{3x - 4}{5} = 7 \;\Rightarrow\; 3x - 4 = 35 \;\Rightarrow\; 3x = 39 \;\Rightarrow\; x = 13

Clear fractions early by multiplying both sides by a common denominator, and expand brackets before collecting terms. When the variable appears on both sides, gather it on the side where its coefficient stays positive; fewer negative signs means fewer slips.

Two variables: elimination or substitution

A single equation in two unknowns, such as 2x+3y=122x + 3y = 12, has infinitely many solutions: every point of a line. A second, independent equation pins down one point.

  • Elimination. Scale one or both equations so that one variable has opposite coefficients, then add. It is usually fastest when coefficients are already matched or are small multiples of each other.
  • Substitution. Solve one equation for a variable and put the expression into the other. It is fastest when one variable already has coefficient 1, as in x−y=6x - y = 6.

Whichever you use, check the pair in the equation you did not use last. That costs ten seconds and catches most arithmetic errors.

Aim at the quantity asked for

Many questions ask for a combination, not for xx and yy separately. Before solving, look at what adding or subtracting the equations produces.

If 4a+7b=304a + 7b = 30 and 7a+4b=367a + 4b = 36, adding gives 11a+11b=6611a + 11b = 66, so a+b=6a + b = 6 in one line. Subtracting the first from the second gives 3a−3b=63a - 3b = 6, so a−b=2a - b = 2. If the question wanted a+ba + b, you are already done; solving fully would have taken three times as long.

The habit is simple: read the last line of the question first, then ask whether some sum, difference or multiple of the given equations is that quantity, or a multiple of it.

How many solutions?

Two linear equations in two unknowns behave like two lines:

Situation Lines Solutions
Coefficients not in proportion cross once exactly one
Coefficients in proportion, constants not parallel none
Everything in proportion the same line infinitely many

So 3x−4y=53x - 4y = 5 and 6x−8y=76x - 8y = 7 have no solution: the left sides are in ratio 1:21:2, but the right sides are not. And x+2y=4x + 2y = 4 with 3x+6y=123x + 6y = 12 is really one equation written twice. Two equations do not always mean one answer; that idea returns in data sufficiency.

Words into equations

Most linear questions arrive as a story. Name each unknown with a letter that reminds you what it counts (aa for adult tickets, cc for child tickets), write one equation per fact, and check the units: a count equation adds counts, a money equation adds price times count. When the story has two kinds of thing, a total number and a total value, it is a two-variable system; substituting the count equation into the value equation is usually quickest.

Worked examples

Example 1: Elimination when the coefficients already match

If 3x+2y=233x + 2y = 23 and 5x−2y=175x - 2y = 17, what is the value of x−yx - y?

  1. The yy coefficients are +2+2 and −2-2, so adding the equations removes yy at once: 8x=408x = 40, and x=5x = 5.

  2. Substitute into the first equation: 15+2y=2315 + 2y = 23, so 2y=82y = 8 and y=4y = 4.

  3. Check in the equation not used last: 5(5)−2(4)=25−8=175(5) - 2(4) = 25 - 8 = 17. Correct.

  4. The question asks for x−y=5−4=1x - y = 5 - 4 = 1.

Answer

x−y=1x - y = 1. Stopping at x=5x = 5 and choosing it is the common slip: read the last line again before you answer.

Example 2: Going straight to a+ba + b

If 4a+7b=304a + 7b = 30 and 7a+4b=367a + 4b = 36, what is a+ba + b?

  1. The coefficients are swapped between the equations, a sign that adding them will give equal coefficients.

  2. Add: 11a+11b=6611a + 11b = 66. Divide by 11: a+b=6a + b = 6.

  3. Second route, as a check: subtracting gives 3a−3b=63a - 3b = 6, so a−b=2a - b = 2. With a+b=6a + b = 6 this gives a=4a = 4, b=2b = 2, and indeed 4(4)+7(2)=304(4) + 7(2) = 30 and 7(4)+4(2)=367(4) + 4(2) = 36.

Answer

a+b=6a + b = 6, found in one line. Solving for aa and bb separately also works but takes longer.

Example 3: A count equation and a value equation

A café sold 40 drinks, all coffees at $3\$3 each or teas at $2\$2 each, for a total of $104\$104. How many coffees did it sell?

  1. Let cc be coffees and tt teas. Count: c+t=40c + t = 40. Value: 3c+2t=1043c + 2t = 104.

  2. Substitute t=40−ct = 40 - c: 3c+80−2c=1043c + 80 - 2c = 104, so c=24c = 24 and t=16t = 16.

  3. Check the value: 3(24)+2(16)=72+32=1043(24) + 2(16) = 72 + 32 = 104. Correct.

  4. Quick second route: if all 40 were teas the total would be $80\$80; each coffee instead of a tea adds $1\$1, and 104−80=24104 - 80 = 24 coffees.

Answer

24 coffees. Answering 16 means you solved for the teas.

Common traps

  • Solving for everything when a combination was asked

    If the question asks for x+yx + y, 2x−y2x - y or similar, test whether adding, subtracting or scaling the given equations produces it. Solving for each variable is never wrong, but it costs time and gives more chances for an arithmetic slip.

  • Subtracting an equation but not every term

    When you subtract (x−y=6)(x - y = 6) from another equation, the −y-y becomes +y+y. Write the subtraction in brackets, (2x+5y)−(x−y)(2x + 5y) - (x - y), and expand it before simplifying.

  • Assuming two equations always give one answer

    Two equations whose coefficients are in proportion describe parallel lines (no solution) or the same line (infinitely many). Before solving, glance at the ratios of the coefficients.

  • Answering with the wrong variable

    Word problems define two unknowns and ask for one. After solving, reread the question and label your result ("24 coffees") before you match it to a choice.

Check yourself

Answer each question before you open its answer.

Question 1: Combine, do not solve

Problem Solving: algebra

If 2x+5y=262x + 5y = 26 and x−y=6x - y = 6, what is the value of 3x+4y3x + 4y?

  1. A.20
  2. B.24
  3. C.26
  4. D.32
  5. E.38
Show the answer

Answer: D. 32

Working

Adding the two equations gives (2x+5y)+(x−y)=3x+4y=26+6=32(2x + 5y) + (x - y) = 3x + 4y = 26 + 6 = 32. Check by solving: x=y+6x = y + 6, so 2y+12+5y=262y + 12 + 5y = 26, y=2y = 2 and x=8x = 8; then 3(8)+4(2)=24+8=323(8) + 4(2) = 24 + 8 = 32.

Why the other options are wrong

  • A. This is x+6yx + 6y, what you get by subtracting the second equation from the first instead of adding them.
  • B. This is 3x3x alone with x=8x = 8; the 4y4y term was left out.
  • C. This is the value of the first equation, 2x+5y2x + 5y, not the expression asked for.
  • E. This is 3y+4x3y + 4x: the values x=8x = 8 and y=2y = 2 were swapped when substituting.

What carries over

Read what is asked, then look for a sum or difference of the equations that produces it.

Question 2: When does the system fail?

Problem Solving: algebra

For which value of kk does the system 3x−ky=53x - ky = 5 and 6x−8y=76x - 8y = 7 have no solution?

  1. A.−4-4
  2. B.44
  3. C.88
  4. D.1616
  5. E.No value of kk
Show the answer

Answer: B. 44

Working

No solution means the coefficients are in proportion but the constants are not. The xx coefficients are in ratio 3:6=1:23:6 = 1:2, so the yy coefficients need −k:−8=1:2-k : -8 = 1:2, giving k=4k = 4. Then doubling the first equation gives 6x−8y=106x - 8y = 10, which cannot also equal 7: parallel, distinct lines. Any other kk gives lines that cross once.

Why the other options are wrong

  • A. With k=−4k = -4 the first equation is 3x+4y=53x + 4y = 5, whose coefficients are not in proportion with 6x−8y6x - 8y; the lines cross once.
  • C. This copies the yy coefficient of the second equation without scaling it; 3x−8y=53x - 8y = 5 and 6x−8y=76x - 8y = 7 still meet at one point.
  • D. This scales the wrong way: the second equation is double the first, so kk is half of 8, not double it.
  • E. A system fails exactly when the lines are parallel and distinct, and one value of kk makes them so.

What carries over

Proportional coefficients with non-proportional constants mean parallel lines and no solution.

Question 3: Adult and child tickets

Problem Solving: algebra

A theatre sold 180 tickets. Adult tickets cost $12\$12 and child tickets $7\$7, and ticket sales totalled $1,810\$1{,}810. How many adult tickets were sold?

  1. A.50
  2. B.70
  3. C.90
  4. D.110
  5. E.150
Show the answer

Answer: D. 110

Working

Let aa be adult and cc child tickets: a+c=180a + c = 180 and 12a+7c=181012a + 7c = 1810. Substitute c=180−ac = 180 - a: 12a+1260−7a=181012a + 1260 - 7a = 1810, so 5a=5505a = 550 and a=110a = 110, c=70c = 70. Check: 12(110)+7(70)=1320+490=181012(110) + 7(70) = 1320 + 490 = 1810. Second route: if all 180 were child tickets, sales would be $1,260\$1{,}260; each adult ticket adds $5\$5, and 5505=110\frac{550}{5} = 110.

Why the other options are wrong

  • A. This comes from substituting c=180−ac = 180 - a but multiplying only the 180 by 7: 12a+1260−a=181012a + 1260 - a = 1810.
  • B. This is the number of child tickets, or the result of attaching $12 to child tickets and $7 to adults.
  • C. This assumes half the tickets were adult tickets; at that split sales would be $1,710, not $1,810.
  • E. This divides $1,810 by $12 (about 150.8) and rounds down, as if every ticket were an adult ticket, ignoring the child tickets.

What carries over

In a count-and-value problem, substitute the count equation into the value equation, then label the result.

Key points

  • Undo operations in reverse order, and clear fractions early.

  • Use elimination when coefficients match, substitution when a variable has coefficient 1.

  • Before solving a system, check whether adding or subtracting the equations gives the asked quantity directly.

  • Proportional coefficients mean no solution or infinitely many, never exactly one.

All questions here are original and written for UniExamer; none comes from the test maker.

Written by the UniExamer team with AI assistance. Every worked example and check question was re-solved by an independent checker before publication. No subject expert has reviewed this lesson yet. If you find a mistake, write to support@uniexamer.com.

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