Chapter 6.3 — Exercise 6.2 — Solutions

Checking a Candidate the Same Way Every Time

Deciding whether any given pair of numbers is a genuine solution never needs anything more elaborate than direct substitution followed by a straightforward comparison of the two resulting sides. For 4x − 3y = 6, testing (2, 3): LHS = 4(2) − 3(3) = 8 − 9 = −1, but RHS = 6. Since LHS ≠ RHS, (2, 3) fails — it is not a solution, even though it looked like a reasonable guess. Nothing about how close a wrong candidate comes matters here — LHS = −1 is not particularly far from RHS = 6 in absolute terms, but "close" carries no meaning for an equation; either the two sides come out exactly equal, or the pair is rejected outright, with no partial credit for a near miss.

Finding Three Solutions, Systematically

Rather than guessing at coordinate pairs and checking each one by luck, Question 1 fixes one variable at a convenient value — usually 0 or 1, whichever keeps the remaining arithmetic simplest — and solves the equation for whatever the other variable must then be.

Every row confirms the same underlying fact: fixing either variable at any chosen value always leaves a simple one-variable equation for the other, solvable by exactly the same one-variable methods that have been familiar since much earlier classes, long before two-variable equations were ever introduced. Different choices of the fixed value simply produce different, but equally valid, solutions each time — there's nothing special about 0 or 1 beyond how little arithmetic they demand. Notice too that the second column is empty for y = 6x and x + y = 0 — both equations already pass through (0, 0), so fixing y = 0 lands right back on the same point already found by fixing x = 0, rather than turning up anything new. That's not a coincidence specific to these two: any equation with no constant term, where c = 0 in the general form, is guaranteed to have (0, 0) as one of its solutions, since substituting zero for both variables leaves every term equal to zero.

Working Backward From a Known Point

Question 2 reverses the direction: given that (0, a) and (b, 0) both satisfy an equation, find a and b. For the equation 8x − y = 34: substituting the point (0, a) directly in gives −a = 34, so a = −34; substituting the point (b, 0) into the same equation gives 8b = 34, so b = 17/4. Both substitutions use the exact same equation, just with a different coordinate already fixed at zero each time. This is really the same technique from Question 1 run in reverse: instead of choosing a convenient value and solving for the unknown coordinate, the point's structure — one coordinate already known to be zero — is chosen for you, and the job is only to solve the resulting one-variable equation for whichever letter, a or b, is standing in for the missing coordinate.

Solving for an Unknown Coefficient

Questions 4 and 6 flip the usual direction again: instead of finding a solution, a solution is given and an unknown constant inside the equation itself must be found. If x = 2, y = 1 satisfies 2x + 3y = k, substituting gives 2(2) + 3(1) = k, so k = 7 — and the now-fully-known equation 2x + 3y = 7 can then be solved for further solutions exactly as in Question 1 — finding k first is really just clearing away the one piece of missing information standing between "an equation with an unknown constant in it" and "an ordinary equation ready for the same three-solutions treatment already practiced." The same idea handles two unknowns tangled together: if x = 2 − α and y = 2 + α satisfies 3x − 2y + 6 = 0, substituting and simplifying gives 8 − 5α = 0, so α = 8/5. What makes this second example harder than the first isn't any new technique — it's that x and y are both written in terms of the single unknown α before substitution even begins, so the equation being solved for α only emerges after both expressions have been plugged into the original 3x − 2y + 6 = 0 and the two variables have fully collapsed into one.

Testing Several Candidates Against One Equation

Question 3 checks several candidate points against 2x − 5y = 10 in one pass: (0, 2) fails (LHS = −10), (0, −2) works (LHS = 10), (5, 0) works (LHS = 10), and (1/2, 2) fails (LHS = 1 − 10 = −9). Only two of the four genuinely satisfy the equation — a reminder that being a solution is never simply a matter of a candidate point merely looking plausible at a glance, only of the arithmetic actually working out equal on both sides once everything is substituted in and fully simplified. Testing several candidates back to back like this, rather than just one in isolation, also makes a useful pattern visible: the points that do work, (0, −2) and (5, 0), share nothing about their appearance that would have marked them out in advance — one has a zero abscissa, the other a zero ordinate, and only carrying out the substitution itself ever reveals which candidates genuinely belong to the line and which ones merely resemble solutions at first glance without actually being any.

From Algebra to a Picture

Every single solution found across the whole of this exercise is really just a point that could be marked directly on the Cartesian plane from Chapter 5. Exercise 6.3 does exactly that with several solutions of the same equation at once, revealing what the infinitely many solutions of a single linear equation actually look like together: not a scatter of unrelated points, but one continuous straight line. Every technique built here — substituting to check a candidate, fixing one coordinate to find another, solving backward for an unknown constant — carries over unchanged; only the underlying goal shifts, moving from finding individual solutions one at a time toward seeing the entire, infinite solution set of the equation all at once, in a single unbroken picture.