Linear equations - CENTS INSIGHT

Linear equations are among the most important topics in algebra and a fundamental part of SAT and aptitude test preparation. They help you solve problems involving unknown values, understand relationships between quantities, and translate written situations into mathematical expressions.

In aptitude tests, linear equations often appear in questions about numbers, ages, prices, ratios, distance, and everyday problem-solving. In the SAT, they are essential for the Math section and provide the foundation for more advanced topics, including systems of equations, linear inequalities, and functions. Questions may test more than your ability to solve an equation; they may require you to form an equation from a word problem, identify an unknown value, or determine how changing one quantity affects another.

Before You Start: Expressions vs. Linear Equations

Before learning how to solve linear equations, it is important to understand the difference between an algebraic expression and an equation.

1. What Is an Algebraic Expression?

An algebraic expression is a combination of numbers, variables, and mathematical operations such as addition, subtraction, multiplication, and division. It does not contain an equals sign.

Examples:

3x + 5
2x − 7
4a + 3b
x² + 2x + 1

2. What Is a Linear Equation?

A linear equation is a mathematical statement that says two expressions are equal. It contains an equals sign (=), and in a linear equation, the variable has a power of 1.

Examples:

3x + 5 = 20
2x − 7 = 9
4x + 3 = 19

Quick Quiz: Test Yourself

Which of the following is not a linear equation?

Show Solution

Correct answer: Option 4, 8x + 7xy = 10.

This question is about linear equations. A linear equation has variables only to the first power, and variables must not be multiplied by one another.

In Option 4, the term 7xy contains the product of two variables, x and y. This makes the equation nonlinear.

Why the other options are linear:

Option 1: 4x + 10 = 0 is linear because x has power 1.
Option 2: 2x + 3² = 0 simplifies to 2x + 9 = 0. The exponent applies to the constant 3, not to x, so the equation remains linear.
Option 3: 8x + 3y = 7 is linear because both variables have power 1 and are not multiplied together.

Characteristics of a linear equation:

Each variable has a power of exactly 1 in every variable term.
Variables are not multiplied or divided by other variables.
Variables do not appear inside roots or as exponents.
Coefficients are constants.
A linear equation in one variable can be written as ax + b = 0, where a is not zero.
A linear equation in two variables can be written as ax + by = c, where a and b are not both zero.

Important distinction: All four choices are equations because they contain an equals sign. Option 4 is nonlinear; it is still an equation.

Solving One-Variable Linear Equations

A linear equation in one variable can be written as ax + b = c where a ≠ 0. The variable has power 1 only. “Solving” means finding the value that makes both sides equal. The method is always the same: do identical operations to both sides until the variable stands alone.

Step-by-step routine

Clear fractions by multiplying every term by the LCM of the denominators.
Expand brackets using the distributive law: a(b + c) = ab + ac.
Combine like terms on each side separately.
Collect variable terms on one side and constants on the other.
Divide by the coefficient of the variable, then check by substituting back.

Three possible outcomes

Result after simplifying	Meaning	Example
x = number	Exactly one solution	2x = 8 gives x = 4
False statement (5 = 9)	No solution (contradiction)	x + 5 = x + 9
True statement (7 = 7)	Infinitely many solutions (identity)	2(x + 1) = 2x + 2

Understanding the concept

Think of an equation as a balance scale: both pans hold the same weight. Whatever you add, subtract, multiply or divide on one side, you must do on the other, or the scale tips. Solving is simply a sequence of balanced moves that “peel away” everything around x, working in reverse order of operations (undo addition/subtraction first, then multiplication/division). In SAT questions the equation is often disguised with brackets or fractions, so the first job is always to simplify it into the plain form ax + b = c. Also notice the special cases: if the x terms cancel completely, the leftover statement tells you whether there is no solution or infinitely many.

Example 1Solve 3(2x − 5) − 4 = 2(x + 3) + 1

Step 1: Expand both sides. Multiply each bracket term by the number outside.

6x − 15 − 4 = 2x + 6 + 1

Step 2: Combine like terms. Constants on the left: −15 − 4 = −19. On the right: 6 + 1 = 7.

6x − 19 = 2x + 7

Step 3: Move variables left, constants right. Subtract 2x and add 19 to both sides.

4x = 26

Step 4: Divide by 4.

x = 26/4 = 13/2 = 6.5

Check: Left = 3(13 − 5) − 4 = 24 − 4 = 20. Right = 2(9.5) + 1 = 20. ✓

Example 2Solve (x + 1)/3 + (x − 2)/6 = 4

Step 1: Find the LCM of 3 and 6, which is 6. Multiply every term by 6 (including the 4 on the right).

2(x + 1) + (x − 2) = 24

Step 2: Expand and combine.

2x + 2 + x − 2 = 24 3x = 24

Step 3: Divide by 3. x = 8.

Check: 9/3 + 6/6 = 3 + 1 = 4. ✓

Exam tip: Forgetting to multiply the whole-number term by the LCM is the most common error here.

Example 3Solve 7x − 3(x − 4) = 2(x + 6) + 8

Step 1: Expand. Note −3 multiplies both x and −4, giving −3x + 12.

7x − 3x + 12 = 2x + 12 + 8

Step 2: Combine.

4x + 12 = 2x + 20

Step 3: Isolate. Subtract 2x and 12: 2x = 8, so x = 4.

Check: Left = 28 − 3(0) = 28. Right = 2(10) + 8 = 28. ✓

Example 4For what value of c does 4(2x − 3) = 8x + c have infinitely many solutions? What happens for any other c?

Step 1: Expand the left side.

8x − 12 = 8x + c

Step 2: Subtracting 8x from both sides removes x entirely, leaving −12 = c.

Infinitely many solutions occur when this statement is true, so c = −12. For any other c the statement (such as −12 = 5) is false, so there is no solution. The SAT often tests this: when x cancels, the constants decide the outcome.

Slope, Intercepts and Forms of a Line

A linear equation in two variables graphs as a straight line. Three numbers describe it: the slope (steepness), the y-intercept (where it crosses the y-axis, at x = 0) and the x-intercept (where it crosses the x-axis, at y = 0).

Slope: m = (y₂ − y₁)/(x₂ − x₁) = rise/run. Positive slope rises left to right, negative falls, zero is horizontal, and a vertical line has undefined slope.

Form	Equation	Best used when
Slope-intercept	y = mx + b	You know the slope and y-intercept
Point-slope	y − y₁ = m(x − x₁)	You know the slope and one point
Standard	Ax + By = C	Finding intercepts, solving systems

From standard form: slope = −A/B, x-intercept = C/A, y-intercept = C/B.

Understanding the concept

Slope measures how fast y changes when x increases by 1. A slope of 3/2 means “up 3 for every 2 steps right”. The y-intercept is the starting height of the line, and the x-intercept is where the output becomes zero, which often answers questions like “when does it run out?”. The three forms are the same line written differently: use slope-intercept to read the rate and start, point-slope to build a line quickly from a point and slope, and standard form to find intercepts or to solve systems. Being able to switch between forms quickly is a key SAT skill.

Example 1Find the equation of the line through (−2, 5) and (4, −1), and its intercepts.

Step 1: Slope.

m = (−1 − 5)/(4 − (−2)) = −6/6 = −1

Step 2: Point-slope using (−2, 5).

y − 5 = −1(x + 2) y = −x − 2 + 5 y = −x + 3

Step 3: Intercepts. y-intercept is 3 (the constant). For the x-intercept set y = 0: 0 = −x + 3, so x = 3.

Check with the other point: −4 + 3 = −1. ✓

Example 2For 6x − 4y = 24, find the slope and both intercepts.

Method A, formulas: slope = −A/B = −6/(−4) = 3/2. x-intercept = 24/6 = 4. y-intercept = 24/(−4) = −6.

Method B, rearranging to confirm:

−4y = −6x + 24 y = (3/2)x − 6

Slope 3/2 and y-intercept −6 agree. For the x-intercept: 0 = 1.5x − 6, so x = 4. ✓

Example 3A line has x-intercept 6 and y-intercept −3. Write its equation in slope-intercept and standard form.

Step 1: The intercepts are the points (6, 0) and (0, −3). Slope:

m = (−3 − 0)/(0 − 6) = 1/2

Step 2: The y-intercept is −3, so y = (1/2)x − 3.

Step 3: Standard form. Multiply by 2: 2y = x − 6, so x − 2y = 6.

Check: (6, 0) gives 6 − 0 = 6 ✓.

Example 4A line through (1, 2) has slope −3. Where does it cross the x-axis?

Step 1: Point-slope form.

y − 2 = −3(x − 1) y = −3x + 3 + 2 = −3x + 5

Step 2: The x-axis crossing has y = 0: 0 = −3x + 5, so x = 5/3.

The point is (5/3, 0). Tip: setting the other variable to 0 is the standard way to find any intercept.

Modeling Real Situations (SAT Word Problems)

SAT questions often describe a constant rate and ask you to build or interpret y = mx + b. Translate the words like this:

m (slope) is the quantity changing “per”, “each”, “every” or “at a rate of”.
b (intercept) is the starting amount, fixed fee or initial value (the value when x = 0).
A negative slope means the quantity decreases (draining, depreciating, burning).

Interpretation questions ask “what does the 2.5 represent?” Always answer with units: “the cost in dollars for each additional kilometer”.

Understanding the concept

A linear model describes anything that changes at a constant rate. The equation y = mx + b is a sentence in disguise: “start with b, then add m for every unit of x”. To build a model, find the starting value (b) and the rate (m); to interpret one, say what m and b mean in context, including units. If you are given two data points instead, compute the slope first, then substitute one point to find b. Questions that compare two plans (two phone plans, two gyms) are solved by setting the two models equal to find the break-even point.

Example 1A taxi charges a fixed $4 plus $2.50 per kilometer. Write the cost model, then find the distance for a $29 fare.

Model: fixed fee is b = 4 and the rate is m = 2.5, so C = 2.5d + 4.

Solve 2.5d + 4 = 29:

2.5d = 25 d = 10 km

Interpretation: 2.5 is the cost per kilometer; 4 is the cost of a 0 km trip (the booking fee).

Example 2A tank holds 80 L. It drains at a constant rate, and after 6 minutes it holds 50 L. When is it empty?

Step 1: Two data points are (0, 80) and (6, 50). Slope:

m = (50 − 80)/(6 − 0) = −5 L per minute

Step 2: Model V = −5t + 80.

Step 3: Empty means V = 0.

0 = −5t + 80 t = 16 minutes

The negative slope shows the tank is losing 5 L every minute.

Example 3Gym A charges $30 to join plus $15 per month. Gym B has no joining fee and charges $25 per month. After how many months do the two cost the same?

Step 1: Models for total cost after m months: A = 15m + 30 and B = 25m.

Step 2: Set equal (break-even).

15m + 30 = 25m 30 = 10m m = 3

Step 3: Interpret. At 3 months both cost $75. After that, B grows faster (slope 25 vs 15), so A is cheaper in the long run.

Example 4A plant’s height in cm is modeled by h = 12 + 0.8w, where w is weeks. (a) What does 0.8 mean? (b) When does it reach 40 cm?

(a) 0.8 is the slope: the plant grows 0.8 cm each week. The 12 is its initial height at w = 0.

(b) Solve 12 + 0.8w = 40:

0.8w = 28 w = 35 weeks

A good sanity check: 35 weeks × 0.8 = 28 cm of growth, plus 12 cm start, equals 40 cm. ✓

Solving Systems: Substitution and Elimination

A system is two or more equations solved together. The solution is the ordered pair (x, y) that satisfies both, which is the point where the two lines cross.

Method	Use it when	Idea
Substitution	One variable is already isolated (y = … or x = …)	Replace that variable in the other equation
Elimination	Both equations are in Ax + By = C form	Multiply to match coefficients, then add or subtract to cancel one variable

Understanding the concept

Each equation in a system is a line, and the solution is the single point lying on both. Both methods transform the system into one equation in one variable. Substitution works by rewriting one variable in terms of the other and plugging it in. Elimination works by scaling the equations so one variable has opposite coefficients; adding the equations then cancels it. Choose elimination when the equations are in Ax + By = C form (no fractions to create), and substitution when one variable already has coefficient 1 or is isolated. Always find the second variable by back-substitution and check in both equations.

Example 1Substitution: y = 2x − 3 and 3x + 2y = 22

Step 1: y is isolated, so replace y in the second equation with (2x − 3).

3x + 2(2x − 3) = 22 3x + 4x − 6 = 22 7x = 28 x = 4

Step 2: Back-substitute: y = 2(4) − 3 = 5.

Solution: (4, 5). Check in the second equation: 12 + 10 = 22. ✓

Example 2Elimination: 3x + 4y = 18 and 5x − 2y = 4

Step 1: Make the y-coefficients opposites. Multiply the second equation by 2:

10x − 4y = 8

Step 2: Add it to the first equation. The y terms cancel.

3x + 4y = 18 10x − 4y = 8 ———– 13x = 26 → x = 2

Step 3: 3(2) + 4y = 18 gives 4y = 12, so y = 3.

Solution: (2, 3). Check: 5(2) − 2(3) = 10 − 6 = 4. ✓

Example 3Solve x − 3y = 7 and 2x + 5y = 3.

Step 1: The first equation gives x easily: x = 7 + 3y. Substitute into the second.

2(7 + 3y) + 5y = 3 14 + 6y + 5y = 3 11y = −11 → y = −1

Step 2: x = 7 + 3(−1) = 4.

Solution: (4, −1). Check: 2(4) + 5(−1) = 8 − 5 = 3 ✓.

Example 4Solve 4x + 3y = 1 and 3x + 2y = 0.

Step 1: Neither variable cancels directly, so scale both equations to make the y-coefficients equal (LCM of 3 and 2 is 6). Multiply the first by 2 and the second by 3.

8x + 6y = 2 9x + 6y = 0

Step 2: Subtract the first from the second: x = −2.

Step 3: 3(−2) + 2y = 0, so y = 3.

Solution: (−2, 3). Check: 4(−2) + 3(3) = −8 + 9 = 1 ✓.

Word Problems That Become Systems

When a problem has two unknowns and two pieces of information, define a variable for each unknown and write one equation per fact. Typical SAT and aptitude patterns are:

Count equation: total items = a + c.
Value equation: total money = (price × number) + (price × number).
Mixtures: amount of pure substance = percent × volume, added over both solutions.

Understanding the concept

Word problems become systems when two quantities are unknown. The secret is a reliable translation routine: (1) name the unknowns with clear letters, (2) underline each separate fact in the problem, (3) turn each fact into one equation, (4) solve with substitution or elimination, (5) reread the question and answer exactly what it asks. Most facts fall into a few types: totals (“altogether”), values (price times quantity), relationships (“twice as many”, “6 more than”), and rates (distance = speed times time). Count your equations: you need as many as you have unknowns.

Example 1Adult tickets cost $12 and child tickets $7. A total of 50 tickets brought in $475. How many of each were sold?

Define: a = adult tickets, c = child tickets.

a + c = 50 12a + 7c = 475

Substitute c = 50 − a into the second equation:

12a + 7(50 − a) = 475 12a + 350 − 7a = 475 5a = 125 → a = 25

Then c = 25. Check: 12(25) + 7(25) = 300 + 175 = 475. ✓

Example 2How many litres of a 10% acid solution and a 30% acid solution must be mixed to make 20 L of 22% solution?

Define: x = litres of 10%, y = litres of 30%.

x + y = 20 (volume) 0.10x + 0.30y = 0.22(20) = 4.4 (pure acid)

Substitute x = 20 − y:

0.10(20 − y) + 0.30y = 4.4 2 − 0.1y + 0.3y = 4.4 0.2y = 2.4 → y = 12

So y = 12 L of 30% and x = 8 L of 10%. Sense check: 22% is closer to 30% than 10%, so more of the 30% solution is needed. ✓

Example 3Ali is 6 years older than twice Sara’s age. Together their ages total 36. How old is each?

Define: a = Ali’s age, s = Sara’s age. The two facts give:

a = 2s + 6 a + s = 36

Substitute: (2s + 6) + s = 36, so 3s = 30 and s = 10. Then a = 26.

Check: 2(10) + 6 = 26 ✓ and 26 + 10 = 36 ✓.

Example 4A boat travels 30 km downstream in 2 hours and returns the same distance upstream in 3 hours. Find the boat’s speed in still water and the current’s speed.

Define: b = boat speed, c = current speed. Downstream the current helps; upstream it hinders. Speed = distance ÷ time.

b + c = 30/2 = 15 b − c = 30/3 = 10

Add the equations: 2b = 25, so b = 12.5 km/h. Then c = 15 − 12.5 = 2.5 km/h.

Check: 12.5 + 2.5 = 15 ✓ and 12.5 − 2.5 = 10 ✓.

Number of Solutions of a System

Two lines can intersect once, never, or coincide. Compare the equations in the form a₁x + b₁y = c₁ and a₂x + b₂y = c₂:

Condition	Lines	Solutions
a₁/a₂ ≠ b₁/b₂ (different slopes)	Intersect	Exactly one
a₁/a₂ = b₁/b₂ ≠ c₁/c₂	Parallel	None
a₁/a₂ = b₁/b₂ = c₁/c₂	Same line	Infinitely many

SAT favorite: “For what value of k does the system have no solution?” Make the slopes equal, then confirm the constants are not in the same ratio.

Understanding the concept

Geometrically, two lines in a plane either cross once, never meet (parallel), or lie on top of each other. Algebraically, this is decided by slopes and intercepts. If slopes differ, there is exactly one intersection. If slopes are equal but intercepts differ, the lines never meet. If both are equal, every point is shared. When you solve by substitution or elimination, the outcome shows up directly: a variable value means one solution, a false statement like 0 = 5 means none, and a true statement like 0 = 0 means infinitely many.

Example 1For what value of k does 2x + 3y = 7 and kx + 6y = 20 have no solution?

Step 1: Equal slopes means the x and y coefficients are proportional.

2/k = 3/6 = 1/2 → k = 4

Step 2: Check constants. Multiplying the first equation by 2 gives 4x + 6y = 14, but the second says 4x + 6y = 20. Since 14 ≠ 20, the lines are parallel and distinct.

Answer: k = 4.

Example 2The system 3x − y = 5 and 6x − 2y = c has infinitely many solutions. Find c.

Step 1: Multiply the first equation by 2 to match the x-coefficient:

6x − 2y = 10

Step 2: For the lines to be identical, the second equation must also read 6x − 2y = 10, so c = 10.

Note: If c were any other number, the lines would be parallel and there would be no solution. Algebraically, substituting y = 3x − 5 into the second equation gives 6x − 6x + 10 = c, i.e. 10 = c. A true statement means infinite solutions; a false one means none.

Example 3How many solutions does y = 2x + 1 and 4x − 2y = 6 have?

Step 1: Compare slopes. The first has slope 2. Rewrite the second: −2y = −4x + 6, so y = 2x − 3, also slope 2.

Step 2: Same slope, different intercepts (1 vs −3), so the lines are parallel.

Algebraic confirmation: 4x − 2(2x + 1) = 6 gives −2 = 6, which is false. No solution.

Example 4How many solutions does x + 2y = 8 and 3x − y = 3 have? Find them.

Step 1: Slopes are −1/2 and 3. Different slopes mean the lines cross exactly once: one solution.

Step 2: Find it. From the second, y = 3x − 3. Substitute:

x + 2(3x − 3) = 8 7x − 6 = 8 x = 2, y = 3

Solution: (2, 3).

Parallel and Perpendicular Lines

Parallel lines have equal slopes (m₁ = m₂). Perpendicular lines have slopes that are negative reciprocals (m₁ × m₂ = −1). A horizontal line (slope 0) is perpendicular to a vertical one.

Always get the slope from the given line first (rearrange to y = mx + b if needed), then use point-slope form with the point you are given.

Understanding the concept

Parallel lines never meet because they rise at the same rate, so their slopes are identical. Perpendicular lines meet at a right angle: if one rises steeply, the other must fall gently. That is why their slopes are negative reciprocals (for example 2 and −1/2, or 3/4 and −4/3). A quick check is that the product of the slopes equals −1. In problems, first convert any given equation to find its slope, apply the parallel or perpendicular rule, then write the new line from the given point.

Example 1Find the line through (3, −2) that is parallel to 4x − 2y = 9.

Step 1: Slope of the given line. −2y = −4x + 9 gives y = 2x − 4.5, so m = 2.

Step 2: A parallel line has the same slope, m = 2.

y − (−2) = 2(x − 3) y + 2 = 2x − 6 y = 2x − 8

Check: 2(3) − 8 = −2. ✓

Example 2Find the line through (4, 5) perpendicular to y = −(2/3)x + 1.

Step 1: Given slope is −2/3. The negative reciprocal is +3/2 (flip the fraction, change the sign).

Step 2:

y − 5 = (3/2)(x − 4) y = 1.5x − 6 + 5 y = 1.5x − 1

Check: (−2/3)(3/2) = −1 ✓ and 1.5(4) − 1 = 5 ✓.

Example 3Are the lines 3x + 4y = 8 and 8x − 6y = 5 perpendicular?

Step 1: Slopes. For the first, m₁ = −A/B = −3/4. For the second, m₂ = −8/(−6) = 4/3.

Step 2: Product.

(−3/4)(4/3) = −1

The product is −1, so the slopes are negative reciprocals: yes, the lines are perpendicular.

Example 4For what value of k is the line kx + 3y = 9 perpendicular to y = 2x − 5?

Step 1: The given slope is 2, so the perpendicular slope must be −1/2.

Step 2: The slope of kx + 3y = 9 is −k/3. Set equal:

−k/3 = −1/2 k = 3/2

Check: slope = −(3/2)/3 = −1/2, and 2 × (−1/2) = −1 ✓.

Linear Inequalities

Solve inequalities like equations, with one crucial difference:

When you multiply or divide both sides by a negative number, reverse the inequality sign. Example: −2x < 6 becomes x > −3.

Graphing on a number line: open circle for < or >, closed circle for ≤ or ≥.
Compound inequalities: −5 < 2x − 1 ≤ 9 is solved by applying each step to all three parts: −4 < 2x ≤ 10, so −2 < x ≤ 5.
Word phrases: “at most” is ≤, “at least” is ≥, “fewer than” is <, “more than” is >.

Understanding the concept

An inequality describes a range of values rather than a single value. The solution is solved with the same steps as an equation, with one safeguard: multiplying or dividing by a negative reverses the order of numbers (3 > 2 but −3 < −2), so the sign must flip. The answer is a region on the number line. In word problems, keywords decide the sign (“at most” means ≤, “at least” means ≥), and the endpoint is included only when the sign has “or equal to”. A fast self-check is to test one value inside the solution and one outside.

Example 1Solve −3(x − 2) ≥ 2x + 11 and state the solution set.

−3x + 6 ≥ 2x + 11 −5x ≥ 5

Divide by −5. Because we divide by a negative, the sign flips:

x ≤ −1

Check x = −2 (should work): left = −3(−4) = 12, right = −4 + 11 = 7, and 12 ≥ 7 ✓. Check x = 0 (should fail): left = 6, right = 11, and 6 ≥ 11 is false ✓.

Example 2A phone plan costs $20 per month plus $0.10 per minute. Your budget is at most $35. What is the greatest whole number of minutes you can use?

Inequality: “at most $35” means ≤ 35.

20 + 0.10m ≤ 35 0.10m ≤ 15 m ≤ 150

Answer: 150 minutes. At exactly 150 minutes the bill is $20 + $15 = $35, which meets the budget.

Example 3Solve the compound inequality −7 < 3x + 2 ≤ 11 and list the integer solutions.

Step 1: Subtract 2 from all three parts.

−9 < 3x ≤ 9

Step 2: Divide all parts by 3 (positive, so no flip): −3 < x ≤ 3.

Integers: −2, −1, 0, 1, 2, 3, which is 6 values. −3 is excluded (strict <) and 3 is included (≤).

Example 4A salesperson earns $400 plus 8% commission on sales. What sales amount is needed to earn at least $1,000?

Step 1: Let s = sales in dollars. “At least” means ≥.

400 + 0.08s ≥ 1000 0.08s ≥ 600 s ≥ 7500

Answer: sales of at least $7,500. At exactly $7,500, earnings are 400 + 600 = $1,000, which meets the target.

Practice Questions: Test Yourself

Choose an answer to see instantly whether it is right or wrong. If you pick a wrong option, the correct one is highlighted. Then press Show Solution to review the steps.

1. If 4(2x − 3) − 3(x + 2) = 2x + 9, what is the value of x?

Show Solution

Correct answer: D. 9

Expand the brackets:

8x − 12 − 3x − 6 = 2x + 9

Combine like terms:

5x − 18 = 2x + 9

Subtract 2x from both sides and add 18:

3x = 27

x = 9

2. Solve the equation: (x − 2)/3 + (x + 1)/2 = 4.

Show Solution

Correct answer: A. 5

The denominators are 3 and 2. Multiply every term by 6:

2(x − 2) + 3(x + 1) = 24

Expand:

2x − 4 + 3x + 3 = 24

5x − 1 = 24

5x = 25

x = 5

3. If 0.4(5x − 10) = 0.5(2x + 8), what is x?

Show Solution

Correct answer: B. 8

Distribute the decimal coefficients:

2x − 4 = x + 4

Subtract x from both sides:

x − 4 = 4

Add 4 to both sides:

x = 8

4. If 2x + 3y = 19 and x − y = 2, what is the value of x + y?

Show Solution

Correct answer: D. 8

From x − y = 2, we obtain x = y + 2.

Substitute into the first equation:

2(y + 2) + 3y = 19

2y + 4 + 3y = 19

5y = 15, so y = 3.

Then x = 3 + 2 = 5.

Therefore, x + y = 5 + 3 = 8.

5. How many solutions does the equation 4(x − 2) + 3 = 4x − 5 have?

Show Solution

Correct answer: B. Infinitely many solutions

Expand the left-hand side:

4x − 8 + 3 = 4x − 5

Simplify:

4x − 5 = 4x − 5

Both sides are identical. The equation is true for every real value of x, so it has infinitely many solutions.

Remember: an identity has infinitely many solutions, while a contradiction has no solution.

6. Which statement describes the equation 3(2x + 1) = 6x + 5?

Show Solution

Correct answer: A. It has no solution

Expand the brackets:

6x + 3 = 6x + 5

Subtract 6x from both sides:

3 = 5

This statement is impossible. No value of x can make the equation true, so there is no solution.

7. For which value of k does the equation (k + 2)x − 3 = 5x − 3 have infinitely many solutions?

Show Solution

Correct answer: C. 3

For infinitely many solutions, both sides must be identical for every value of x.

Compare the coefficients of x:

k + 2 = 5

Subtract 2:

k = 3

When k = 3, the equation becomes 5x − 3 = 5x − 3, which is an identity.

8. The sum of two numbers is 42. One number is 6 less than twice the other. What is the larger number?

Show Solution

Correct answer: C. 26

Let one number be x. The other number is 2x − 6.

Their sum is 42:

x + (2x − 6) = 42

3x − 6 = 42

3x = 48

x = 16

The other number is 2(16) − 6 = 26. Therefore, the larger number is 26.

9. A service charges a fixed fee plus the same charge per item. Processing 3 items costs Rs. 29, while processing 5 items costs Rs. 43. What is the total cost of processing 4 items?

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Correct answer: D. Rs. 36

Let f be the fixed fee and c the charge per item.

f + 3c = 29

f + 5c = 43

Subtract the first equation from the second:

2c = 14, so c = 7.

Substitute into the first equation:

f + 21 = 29, so f = 8.

For 4 items, the total is 8 + 4(7) = Rs. 36.

10. Solve: 5 − (2x − 1)/3 = (x + 4)/2.

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Correct answer: C. 20/7

The denominators are 3 and 2. Multiply every term by 6:

30 − 2(2x − 1) = 3(x + 4)

Expand:

30 − 4x + 2 = 3x + 12

32 − 4x = 3x + 12

Subtract 12 and add 4x:

20 = 7x

x = 20/7

11. Forty percent of a number is 12 less than half of that number. What is the number?

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Correct answer: A. 120

Let the number be x.

Forty percent is 0.4x, and half the number is 0.5x.

Translate the statement into an equation:

0.4x = 0.5x − 12

Subtract 0.4x from both sides:

0.1x = 12

x = 12 ÷ 0.1 = 120

12. The system below has more than one solution, where u and v are constants:
−2x + uy = v
x − 2y = 3
What is the value of u/v?

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Correct answer: B. −2/3

For two linear equations to have more than one common solution, they must represent the same line.

Multiply the second equation by −2:

−2x + 4y = −6

Compare this with the first equation:

−2x + uy = v

Therefore, u = 4 and v = −6.

Hence, u/v = 4/(−6) = −2/3.