Solving Systems by Elimination: Free Response

Look down the $y$ column. The coefficients are $+5$ and $-5$, opposites, so adding the two equations sends those terms to zero:

$(2x + 5y) + (3x - 5y) = -7 + 27.$

What survives has a single unknown:

$5x = 20, \qquad x = 4.$

Back-substitute into $2x + 5y = -7$:

$8 + 5y = -7, \qquad 5y = -15, \qquad y = -3.$

The test has to include the equation you did NOT back-substitute into, because that is the one your value of $y$ has never been measured against: $3(4) - 5(-3) = 12 + 15 = 27$, and $2(4) + 5(-3) = 8 - 15 = -7$. Both hold, so the pair is $(4, -3)$.

This time the $x$ column holds $+4x$ in both equations. They are equal, not opposite, so adding would produce $8x$ and remove nothing. Subtracting is what cancels them. Take $4x + 3y = 1$ away from $4x + 7y = 13$, sending the minus sign to every term of the equation being removed:

$(4x + 7y) - (4x + 3y) = 13 - 1, \qquad 4y = 12, \qquad y = 3.$

Back-substitute into $4x + 3y = 1$:

$4x + 9 = 1, \qquad 4x = -8, \qquad x = -2.$

Test both: $4(-2) + 3(3) = -8 + 9 = 1$ and $4(-2) + 7(3) = -8 + 21 = 13$.

Combining the equations does arithmetic on the two coefficients sitting in a column, so the variable leaves exactly when that arithmetic lands on zero. Write the two coefficients as $b_1$ and $b_2$.

Adding produces the coefficient $b_1 + b_2$, so

$b_1 + b_2 = 0 \iff b_2 = -b_1.$

Read left to right, opposite coefficients make adding work. Read right to left, if adding worked then the coefficients summed to zero, so they were opposites and nothing weaker would have done. The test holds both ways.

Subtracting produces $b_1 - b_2$, so

$b_1 - b_2 = 0 \iff b_2 = b_1,$

and by the same two readings, subtracting removes a variable exactly when its coefficients are equal.

That is what decided both systems above. One offered a column of $+5$ and $-5$, opposites, so adding was the move; the other offered a column of $4$ and $4$, equal, so subtracting was.

A system offering neither is not stuck. Multiplying an equation through by a nonzero constant rescales its coefficients without changing which pairs satisfy it, so you can manufacture an opposite pair or an equal pair and then combine.

Compare the $y$ coefficients, $4$ and $2$. The smaller divides the larger, so only one equation has to move. Multiply every term of $3x + 2y = 7$ by $-2$, which turns its $+2y$ into $-4y$, the opposite of the $+4y$ above it:

$-6x - 4y = -14.$

Add that to $5x + 4y = 11$:

$(5x + 4y) + (-6x - 4y) = 11 - 14, \qquad -x = -3, \qquad x = 3.$

Back-substitute into $3x + 2y = 7$:

$9 + 2y = 7, \qquad 2y = -2, \qquad y = -1.$

Test both: $5(3) + 4(-1) = 15 - 4 = 11$ and $3(3) + 2(-1) = 9 - 2 = 7$.

Multiplying by $+2$ instead and then subtracting lands in the same place. The constant is chosen to make the two coefficients opposite or equal, and the operation follows from which of the two you aimed at.

Aim at the $y$ coefficients, $3$ and $-5$. Neither divides the other, so no single multiplier can line them up: scaling one equation alone would leave the other coefficient where it was. Send both up to the least common multiple of $3$ and $5$, which is $15$. Multiply $4x + 3y = -1$ by $5$ and $3x - 5y = 21$ by $3$:

$20x + 15y = -5, \qquad 9x - 15y = 63.$

The $y$ coefficients are now $+15$ and $-15$, opposites, so add:

$29x = 58, \qquad x = 2.$

Back-substitute into $4x + 3y = -1$:

$8 + 3y = -1, \qquad 3y = -9, \qquad y = -3.$

Test in both originals: $4(2) + 3(-3) = 8 - 9 = -1$ and $3(2) - 5(-3) = 6 + 15 = 21$.

Weigh the two columns before touching either. The $x$ coefficients are $6$ and $2$, and $2$ divides $6$, so one multiplication lines them up. The $y$ coefficients are $5$ and $3$, which share no factor larger than $1$, so their least common multiple is $15$ and both equations would have to be rescaled. The $x$ column is the cheaper one.

Multiply every term of $2x + 3y = 4$ by $3$:

$6x + 9y = 12.$

Both equations now carry $6x$, equal rather than opposite, so subtract $6x + 5y = 20$ from it:

$(6x + 9y) - (6x + 5y) = 12 - 20, \qquad 4y = -8, \qquad y = -2.$

Back-substitute into $2x + 3y = 4$:

$2x - 6 = 4, \qquad 2x = 10, \qquad x = 5.$

Test both: $6(5) + 5(-2) = 30 - 10 = 20$ and $2(5) + 3(-2) = 10 - 6 = 4$.

The expensive column is not a wrong column. Scaling by $3$ and by $5$ gives $18x + 15y = 60$ and $10x + 15y = 20$, and subtracting leaves $8x = 40$, so $x = 5$ again. It costs two multiplications and larger numbers to arrive at the same pair.

Both leftovers are statements about numbers, and the two statements say opposite things.

In $3x - 6y = 9$ with $-x + 2y = -3$, multiplying $-x + 2y = -3$ by $3$ gives $-3x + 6y = -9$, and adding:

$(3x - 6y) + (-3x + 6y) = 9 + (-9), \qquad 0 = 0.$

That statement is true, and it is true for every $x$ and every $y$. So the elimination has ruled out nothing at all. The reason is visible in the equations themselves: dividing $3x - 6y = 9$ through by $3$ gives $x - 2y = 3$, and multiplying $-x + 2y = -3$ through by $-1$ gives $x - 2y = 3$ as well. The two are one equation in different clothes, so the system asks one question rather than two, and every pair satisfying $x - 2y = 3$ satisfies both. There are infinitely many, one for each value of $y$, described by $x = 2y + 3$.

The other system is a different animal. Its leftover, $0 = 10$, is false, and no choice of $x$ and $y$ can repair a false numerical statement, so no pair satisfies both equations and that verdict was correct.

What went wrong was treating 'the variables cancelled' as the conclusion. The variables cancelling is only the signal to go and read the numerical statement left behind, and that statement can come out either way.

Rescale each equation until the two left-hand sides are identical, then compare what is on the right.

Dividing every term of $2x - 6y = 8$ by $2$ produces an equation with exactly the same solutions:

$x - 3y = 4.$

Multiplying every term of $-x + 3y = 1$ by $-1$ does the same for that one:

$x - 3y = -1.$

Now suppose some pair $(x_0, y_0)$ satisfied both original equations. It would satisfy both rescaled ones, so the single number $x_0 - 3y_0$ would equal $4$ and also equal $-1$. A number has one value, so no such pair exists.

This is the same fact the elimination reported, said without the machinery. Subtracting the two rescaled equations gives $0 = 5$, and $0 = 5$ is precisely the statement that one quantity was asked to take two different values at once.

One system where the condition holds and the conclusion fails is enough to sink the rule.

Take $x + 4y = 5$ and $2x + 8y = 10$. The second equation's coefficients, $2$ and $8$, are exactly twice the first's, so the student's condition is met. But every term of the second is twice the corresponding term of the first, the constant included, so the second is the first in disguise:

$2x + 8y = 10 \iff x + 4y = 5.$

Eliminating leaves $0 = 0$, and the system has infinitely many solutions. The condition held and the conclusion failed, so the rule as stated is wrong.

What the rule left out is the constants. Suppose the second equation's coefficients are $k$ times the first's, with $k$ not zero, so the left-hand sides are proportional. Two cases remain, and only the constants separate them. If $c_2 = kc_1$, the second equation is $k$ times the first throughout, the two ask the same question, and there are infinitely many solutions. If $c_2 \ne kc_1$, the left-hand sides can be scaled to match while the right-hand sides cannot, and there is no solution. The student's rule names only the second case and misses the first.

Both directions of the corrected rule are worth checking. If the coefficients are proportional, the system is one of those two cases, so it never has exactly one solution. And if the coefficients are NOT proportional, neither case can arise: no scaling makes the left-hand sides match, nothing can cancel both variables at once, and the elimination always leaves a genuine equation in one unknown, which pins that unknown to one value.

Each test has two stages: ask whether the coefficient pairs are proportional, and only then look at the constants.

For $4x - 10y = 6$ with $-6x + 15y = -9$, the coefficients $-6$ and $15$ are $-\tfrac{3}{2}$ times $4$ and $-10$, since $-\tfrac{3}{2}(4) = -6$ and $-\tfrac{3}{2}(-10) = 15$. The constant follows the same factor, because $-\tfrac{3}{2}(6) = -9$. The second equation is the first one scaled, so there are infinitely many solutions.

For $4x - 10y = 6$ with $-6x + 15y = 12$, the coefficients are proportional in exactly the same way, but the constant does not follow: $-\tfrac{3}{2}(6) = -9$, and $-9$ is not $12$. Left-hand sides that can be made identical with right-hand sides that cannot is precisely the case with no solution.

For $4x - 10y = 6$ with $-6x + 14y = -9$, test the proportionality before anything else:

$\frac{-6}{4} = -\frac{3}{2}, \qquad \frac{14}{-10} = -\frac{7}{5}.$

These differ, so no single factor scales one coefficient pair onto the other and the system has exactly one solution. The constants have no say in that conclusion, which is why the $-9$ carried over from the first system changes nothing here.

Aim at the $x$ column. Its coefficients are $3$ and $2$, sharing no factor, so scale both to $6$: multiply $3x + 5y = 2$ by $2$ and $2x - 3y = 14$ by $3$:

$6x + 10y = 4, \qquad 6x - 9y = 42.$

Equal coefficients call for subtraction:

$(6x + 10y) - (6x - 9y) = 4 - 42, \qquad 19y = -38, \qquad y = -2.$

Back-substitute into $3x + 5y = 2$:

$3x - 10 = 2, \qquad 3x = 12, \qquad x = 4.$

Test both: $3(4) + 5(-2) = 12 - 10 = 2$ and $2(4) - 3(-2) = 8 + 6 = 14$. Every number written along the way was a whole number.

Isolate a variable. Every coefficient here is $2$, $3$ or $5$, so whichever variable you pick, a fraction appears at once. Taking $y$ from $2x - 3y = 14$:

$-3y = 14 - 2x, \qquad y = \frac{2x - 14}{3}.$

Substitute that into $3x + 5y = 2$:

$3x + 5\left(\frac{2x - 14}{3}\right) = 2.$

Multiply through by $3$ to clear the denominator, then collect:

$9x + 10x - 70 = 6, \qquad 19x = 76, \qquad x = 4.$

Then $y = \frac{2(4) - 14}{3} = \frac{-6}{3} = -2$, and the pair is $(4, -2)$, the same one elimination produced. The route is not harder in principle, but it carried a denominator from its second line to its fifth, and a denominator dropped anywhere in there loses the whole problem.

Set the two roads side by side on the same system. Elimination took two multiplications, one subtraction and one back-substitution, all in whole numbers. Substitution took one rearrangement, which produced a denominator of $3$, and then a clearing step to remove it. Same pair, more places to slip.

The feature that predicts this is visible before any work is done. The coefficients are $3$, $5$, $2$ and $-3$, and none of them is $1$ or $-1$. Isolating a variable divides by its coefficient, so unless some coefficient is $1$ or $-1$, the isolated expression arrives with a denominator. Elimination scales by multiplying rather than dividing, so it does not create denominators at all.

Now look at $y = 2x - 5$ with $4x + 3y = 5$. One equation already presents $y$ on its own, so substitution costs nothing to set up: replace $y$ in the other equation and solve for $x$:

$4x + 3(2x - 5) = 5.$

Elimination is available here too and is not wrong. Rewriting $y = 2x - 5$ as $-2x + y = -5$ puts the system in standard form, after which you would scale to match a column. It is simply more work than reading off an expression that is already isolated.

The rule of thumb: reach for substitution when a variable already stands alone, or carries a coefficient of $1$ or $-1$ somewhere; reach for elimination when both equations are in standard form with no such coefficient anywhere.

The $y$ coefficients are $+3$ and $-3$, so adding removes $y$:

$(x + 3y) + (2x - 3y) = 7 + 5, \qquad 3x = 12, \qquad x = 4.$

Then $4 + 3y = 7$ gives $y = 1$, and both originals hold: $4 + 3 = 7$ and $8 - 3 = 5$.

Now build the combination. Tripling the first equation and adding the second gives, on the left,

$3(x + 3y) + (2x - 3y) = 3x + 9y + 2x - 3y = 5x + 6y,$

and $3(7) + 5 = 26$ on the right, so the built equation is $5x + 6y = 26$. Test the pair in it: $5(4) + 6(1) = 20 + 6 = 26$, true.

Worth noticing what this combination did NOT do: it removed nothing. It is not a step towards a solution, and the pair satisfies it anyway. That is the first sign that the property at work here has nothing to do with cancelling.

Choose any $x$ you like and solve the combined equation for $y$. Taking $x = -2$:

$5(-2) + 6y = 26, \qquad 6y = 36, \qquad y = 6.$

So $(-2, 6)$ satisfies $5x + 6y = 26$. It satisfies neither original: $-2 + 3(6) = 16$, which is not $7$, and $2(-2) - 3(6) = -22$, which is not $5$. Any $x$ other than $4$ produces another such pair, so there are infinitely many of them.

What this shows is that the relationship runs one way only. Every solution of the system satisfies the combination, as the previous part confirmed, but a pair satisfying the combination need not solve the system. A single combined equation is therefore weaker than the system it was built from: information has been thrown away.

That is exactly why elimination does not replace the system with one equation. It replaces one of the two and keeps the other, so the pair of equations together still pins down the same solutions.

Put the hypothesis into usable form first. Saying that $(x_0, y_0)$ satisfies both equations means the two statements

$a_1x_0 + b_1y_0 = c_1, \qquad a_2x_0 + b_2y_0 = c_2$

are true statements about numbers, not equations still waiting to be solved.

Apply the multiplication property of equality to each: if two quantities are equal, multiplying both by the same number leaves them equal. So

$m(a_1x_0 + b_1y_0) = mc_1, \qquad n(a_2x_0 + b_2y_0) = nc_2,$

whatever numbers $m$ and $n$ are, negative ones and zero included.

Now the addition property of equality: adding equals to equals leaves equals. Adding the left sides of those two true statements and the right sides of them:

$m(a_1x_0 + b_1y_0) + n(a_2x_0 + b_2y_0) = mc_1 + nc_2.$

That is precisely the combined equation evaluated at $(x_0, y_0)$, so the pair satisfies it.

Nothing was assumed about $m$ and $n$ beyond their being numbers. So every combination of the two equations holds at every solution of the system, whether or not the combination happens to cancel anything. The combination built earlier was one instance, with $m = 3$ and $n = 1$.

Suppose $(x_1, y_1)$ satisfies the kept equation and the combination:

$a_1x_1 + b_1y_1 = c_1, \qquad m(a_1x_1 + b_1y_1) + n(a_2x_1 + b_2y_1) = mc_1 + nc_2.$

The first of those, scaled by $m$, says that $m(a_1x_1 + b_1y_1) = mc_1$. Subtract that from the second, which is the addition property used in its subtraction form:

$n(a_2x_1 + b_2y_1) = nc_2.$

Every trace of the kept equation has gone. Now divide both sides by $n$. This is the step that needs $n \ne 0$, and it gives

$a_2x_1 + b_2y_1 = c_2,$

so the pair satisfies the equation that was traded away. Put beside the previous part, this closes the argument: that part sends every solution of the original system into the new one, and this part sends every solution of the new system back. The two systems therefore have exactly the same solutions, which is what makes solving the new one an honest answer to the old one.

When $n = 0$ the argument stops at the division, and it stops for a reason. With $n = 0$ the combination reads $m(a_1x + b_1y) = mc_1$, a multiple of the equation you kept, carrying no information whatever about the one you traded. Take $x + 3y = 7$ with $2x - 3y = 5$ and combine with $m = 3$ and $n = 0$, giving $3x + 9y = 21$. The new system, $x + 3y = 7$ together with $3x + 9y = 21$, is satisfied by $(7, 0)$, since $7 + 0 = 7$ and $21 + 0 = 21$. But $(7, 0)$ fails the discarded equation, because $2(7) - 3(0) = 14$, not $5$. Solutions have been invented, which is what multiplying an equation by zero always risks: it erases the equation instead of rescaling it.