Can there be methods other than elimination and substitution to reduce a pair of linear equations in two variables to a linear equation in one variable? If so, describe them.
Step-by-step solution
Idea: Any step that removes one variable while keeping the equations true will do. Comparing two expressions for the same variable is one such step.
- Comparison method: write both equations as “x = …” (or “y = …”) and equate the right-hand sides, which leaves one variable.1 mark
- Example: 7x − 15y = 2 and x + 2y = 3. Then x = 2 + 15y7 and x = 3 − 2y. Equate: 2 + 15y7 = 3 − 2y ⇒ 2 + 15y = 21 − 14y ⇒ 29y = 19 ⇒ y = 1929, and x = 3 − 3829 = 4929 (same as by substitution).1 mark
- Adding and subtracting (when coefficients are swapped, like 5x + 3y = 21, 3x + 5y = 19): adding gives 8(x + y) = 40, so x + y = 5, and subtracting gives 2(x − y) = 2, so x − y = 1. These give x = 3, y = 2.½ mark
- Another general method is the cross-multiplication formula (met in later classes). The graphical method finds the solution as the point where the two lines meet, but it does not turn the pair into one equation and may be inexact.½ mark
Check: The comparison result (4929, 1929) agrees with the substitution answer for the same pair ✓.
Answer to write in the exam
Comparison method: express the same variable from both equations and equate
e.g. x = 2 + 15y7 and x = 3 − 2y ⇒ 2 + 15y = 21 − 14y ⇒ y = 1929, x = 4929
Adding/subtracting when coefficients are swapped: gives x + y and x − y
∴ Yes, other methods exist (comparison, add–subtract, cross-multiplication)
Common mistakes that cost marks
- Calling the graphical method a way to “reduce to one variable”: it finds the answer by drawing, not by reducing.
- In the comparison method, equating expressions for different variables (x = … with y = …).
How this can come in the exam
Solve y = 2x − 1 and y = −x + 8 by comparing the two expressions for y.
Show answer
2x − 1 = −x + 8 ⇒ 3x = 9 ⇒ x = 3 (1 mark); y = 5 (1 mark).Try one yourself
Solve by comparison: x = 4y + 1 and x = 10 − 2y.
Show answer
4y + 1 = 10 − 2y ⇒ y = 32, x = 7.
More questions like this
- Consider the following pairs of linear equations in two variables.
Pair 1: 2x + y = 6 and x − y = 2
Pair 2: x + y = 4 and 2x + 2y = 8
Pair 3: 3x − 2y = 6 and 6x − 4y = 12
Pair 4: x − 2y = 4 and 2x − 4y = 6
For each pair, prepare a table of values (with at least two ordered pairs). Plot the points on the Cartesian plane. Draw the straight lines. - How do we find solutions from the graph? What can we say about the number of solutions when the two lines (i) intersect at a point, (ii) are parallel, (iii) are coincident?
- Solve the pair of linear equations x + 3y = 6 and 2x − 3y = 12 graphically. Comment on the nature of the solution.
- Determine if the lines x + 2y − 4 = 0 and 2x + 4y − 12 = 0 intersect, coincide or are parallel to each other. Then verify your answer by graphing the equations.
- Can you prove that two lines of equal slope are parallel?
(Hint: Consider the equations of the two lines to be y = mx + d1 and y = mx + d2 )
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