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Linear Equations and Inequalities

Mathematics · WAEC and JAMB · SS2 and SS3

Simultaneous equations appear in Paper 2 nearly every year, usually dressed as a word problem about prices or ages. Inequalities turn up in the objectives and in linear programming style questions, and the sign rule is where most candidates fall.

What you need to know

  • Solving a linear equation means doing the same thing to both sides until the unknown stands alone. Whatever crosses the equals sign changes sign, because you are really subtracting it from both sides.
  • Clear fractions first by multiplying every term by the LCM of the denominators. In x/3 + (x - 1)/4 = 2, multiply through by 12 to get 4x + 3(x - 1) = 24, then 7x = 27 and x = 27/7.
  • In the elimination method, make the coefficients of one letter equal in size, then add if their signs differ and subtract if their signs are the same.
  • In the substitution method, make one letter the subject of the simpler equation and put that expression into the other equation. It is faster when one coefficient is already 1.
  • Graphically, the solution of two simultaneous linear equations is the point where the two lines cross; if the lines are parallel there is no solution, and if they coincide there are infinitely many.
  • For word problems, define your letters in writing first, for example 'let p be the cost of one pen in naira', because the examiner awards a mark for correct formulation before any arithmetic.
  • Always substitute your answers back into the equation you did NOT use to find the last unknown. That one check catches almost every arithmetic slip.
  • An inequality is solved like an equation with one extra rule: multiplying or dividing both sides by a NEGATIVE number reverses the inequality sign. From -3x > 9 you get x < -3, not x > -3.
  • Adding or subtracting a negative number does not reverse the sign; only multiplying or dividing by a negative does.
  • On a number line, an open circle marks a strict inequality (< or >) and a shaded circle marks an inclusive one (<= or >=). Examiners check the circle, not just the direction of the arrow.
  • Combined inequalities are solved in both parts at once: from -5 < 2x - 1 <= 7, add 1 throughout to get -4 < 2x <= 8, then halve to get -2 < x <= 4.
  • When the question asks for the greatest or least INTEGER satisfying an inequality, solve first, then step inward to the nearest whole number. If x <= -4, the greatest integer value is -4; if x < -4, it is -5.
  • The gradient of a straight line through two points is (y2 - y1)/(x2 - x1), and the line is y = mx + c, where c is the intercept on the y-axis. Many 'linear equation' questions are really about reading this from a graph.
  • For a region question, shade or leave clear exactly as the paper instructs, and test one convenient point such as the origin to decide which side of the line satisfies the inequality.

Key terms

Linear equation
An equation in which the highest power of the unknown is one, so its graph is a straight line.
Simultaneous equations
Two or more equations that must be satisfied by the same set of values of the unknowns at the same time.
Elimination method
A method of solving simultaneous equations by adding or subtracting multiples of the equations to remove one unknown.
Inequality
A statement that one quantity is greater than, less than, or not equal to another, usually satisfied by a range of values.
Gradient
The measure of the steepness of a line, equal to the change in y divided by the corresponding change in x.
Solution set
The complete collection of values that satisfy an equation or inequality.

Formulae

  • straight line: y = mx + c, where m is the gradient and c the y-intercept
  • gradient m = (y2 - y1) / (x2 - x1)
  • if a > b then a + c > b + c for any c
  • if a > b and c < 0 then ac < bc (the sign reverses)
  • elimination: multiply each equation so one unknown has equal coefficients, then add or subtract

Worked examples

Solve simultaneously: 3x + 2y = 16 and 5x - 3y = 14.

  1. Choose to eliminate y. Multiply the first equation by 3: 9x + 6y = 48.
  2. Multiply the second equation by 2: 10x - 6y = 28.
  3. The y terms now have opposite signs, so ADD the two new equations: 19x = 76.
  4. Divide by 19: x = 4.
  5. Substitute x = 4 into the first original equation: 3(4) + 2y = 16, so 12 + 2y = 16 and 2y = 4, giving y = 2.
  6. Check in the second original equation: 5(4) - 3(2) = 20 - 6 = 14. Correct.

Three pens and two exercise books cost N1,750. Five pens and four exercise books cost N3,300. Find the cost of one pen and one exercise book.

  1. Let p be the cost of one pen in naira and b the cost of one exercise book in naira.
  2. Form the equations: 3p + 2b = 1750 and 5p + 4b = 3300.
  3. Multiply the first equation by 2 so the b terms match: 6p + 4b = 3500.
  4. Subtract the second equation from this: (6p - 5p) + (4b - 4b) = 3500 - 3300, so p = 200.
  5. Substitute into 3p + 2b = 1750: 600 + 2b = 1750, so 2b = 1150 and b = 575.
  6. Check in the second equation: 5(200) + 4(575) = 1000 + 2300 = 3300. Correct.

Solve the inequality 1 - 2x >= 9 and state the greatest integer value of x that satisfies it.

  1. Subtract 1 from both sides: -2x >= 8.
  2. Divide both sides by -2. Because -2 is negative, the inequality sign REVERSES: x <= -4.
  3. The values allowed are -4 and everything below it.
  4. The greatest of these is -4 itself, since the sign is 'less than or equal to'.
  5. Check: 1 - 2(-4) = 1 + 8 = 9, and 9 >= 9 is true.

The mistake to avoid

Forgetting to reverse the inequality sign after dividing by a negative number turns a correct line of algebra into a wrong solution set, and the follow-on marks go with it. In word problems, candidates also write the equations without ever saying what their letters stand for, and then mix up which answer belongs to which item.

In the exam

State your letters and their units in one line before forming the equations; it is worth a mark and it keeps you from swapping the answers at the end. Always verify in the equation you did not use for the final substitution. If a question says 'illustrate on a number line', draw the line with the correct open or shaded circle, because the diagram carries its own mark.