Solving Systems by Substitution: Free Response

The equation $y = 4 - 2x$ says that $y$ and $4 - 2x$ are two names for one number, so wherever $y$ stands, $4 - 2x$ may stand instead. The place worth doing that is the equation that still has both letters in it:

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

Keep the bracket. The coefficient $2$ multiplies the whole expression that arrived, not just its first term, and dropping the bracket would quietly leave the $-2x$ unmultiplied. Distribute and collect:

$3x + 8 - 4x = 5, \qquad -x + 8 = 5, \qquad -x = -3.$

The letter $y$ has gone, and what is left is an equation of the kind you could already solve before this lesson began.

Divide both sides of $-x = -3$ by $-1$:

$x = 3.$

A number is not yet a solution of a system, because the system asked for a pair. Feed the value back into the equation that was already solved for $y$:

$y = 4 - 2(3) = -2.$

Now test the candidate $(3, -2)$ in the two equations as they were given, not in anything derived along the way:

$4 - 2(3) = -2, \qquad 3(3) + 2(-2) = 9 - 4 = 5.$

Both hold, so $(3, -2)$ is the solution.

Follow the isolation through the work and count the places it does something.

Its first job is the substitution itself. Because $y$ and $4 - 2x$ are the same number, writing one where the other stood turns

$3x + 2y = 5 \quad \text{into} \quad 3x + 2(4 - 2x) = 5,$

and it is this exchange, not any solving, that removes a letter. The equation that gets used up here is $3x + 2y = 5$: after the exchange it has become a statement about $x$ alone.

Its second job comes after the value of $x$ is known. The isolation is still a true statement about the solution, so it converts that value into the other coordinate:

$y = 4 - 2(3) = -2.$

So the two equations are not treated alike. One is transformed into a one-unknown equation and is finished with; the other survives the whole calculation as a rule for producing $y$ from $x$, which is exactly why substituting into it, rather than into its partner, would make no progress at all.

List the four openings and ask what each one would have to divide by.

In $4x + 3y = 18$, isolating $x$ divides by $4$ and isolating $y$ divides by $3$. In $x + 5y = 13$, isolating $y$ divides by $5$. The fourth is different:

$x + 5y = 13 \ \longrightarrow \ x = 13 - 5y.$

The letter $x$ already stands alone in that equation apart from what is added to it, so getting it by itself takes one subtraction and no division at all. That is the feature to scan for: a coefficient of $1$ or $-1$ on some letter in some equation. When one exists, isolating that letter keeps every number in the next line a whole number.

The isolated expression goes into the equation it did not come from:

$4(13 - 5y) + 3y = 18.$

Distribute and collect, and notice that every number on the way is whole:

$52 - 20y + 3y = 18, \qquad -17y = -34, \qquad y = 2.$

Back into the isolation for the other coordinate:

$x = 13 - 5(2) = 3.$

Test $(3, 2)$ in the two equations as given:

$4(3) + 3(2) = 12 + 6 = 18, \qquad 3 + 5(2) = 13.$

Scan the four openings for their divisors first. Isolating $x$ in $2x + 3y = 17$ divides by $2$; isolating $y$ there divides by $3$; isolating $x$ in $5x - 2y = 14$ divides by $5$; isolating $y$ there divides by $2$. So no opening is free, and two of them are tied for the smallest divisor.

Take $y$ from $5x - 2y = 14$:

$2y = 5x - 14, \qquad y = \frac{5x - 14}{2}.$

Substitute into the other equation and clear the denominator at once, by multiplying every term by $2$:

$2x + 3 \cdot \frac{5x - 14}{2} = 17, \qquad 4x + 3(5x - 14) = 34.$

Now the arithmetic is whole again:

$4x + 15x - 42 = 34, \qquad 19x = 76, \qquad x = 4,$

and the isolation returns the other coordinate, $y = \frac{5(4) - 14}{2} = 3$. Test $(4, 3)$ in the equations as given: $2(4) + 3(3) = 17$ and $5(4) - 2(3) = 14$.

The last question is the one worth thinking about. Every opening does the same thing: it writes one quantity in place of another quantity equal to it, which cannot change which pairs make the original two statements true. So the choice sets how heavy the arithmetic is and nothing else. Notice that the answer here is a pair of whole numbers even though the route was full of halves: the fractions were a property of the path, not of the solution.

Read the line as a record of an exchange. The bracket is what took the place of a letter, and the isolation was for $y$, so one of the original equations must have been equivalent to

$y = 2x - 5.$

Put in the requested form, that is $2x - y = 5$.

Everything outside the bracket belonged to the equation that received the expression. Restoring $y$ where the bracket stands, and leaving its other terms untouched, gives

$4x + 3y = 25.$

So the system was $2x - y = 5$ together with $4x + 3y = 25$, and the recorded line is what the second becomes once the first has been used on it.

Expand the bracket and collect:

$4x + 6x - 15 = 25, \qquad 10x = 40, \qquad x = 4.$

The second coordinate comes from the isolation the bracket recorded, not from a fresh calculation:

$y = 2(4) - 5 = 3.$

Test $(4, 3)$ in both recovered equations:

$2(4) - 3 = 5, \qquad 4(4) + 3(3) = 16 + 9 = 25.$

Compare the two equations by asking which pairs satisfy each, which is the only thing an equation is for here.

Every term of $6x - 3y = 15$ is three times the matching term of $2x - y = 5$, so dividing through by $3$ turns one into the other:

$6x - 3y = 15 \ \longleftrightarrow \ 2x - y = 5.$

That step is reversible, since multiplying by $3$ undoes dividing by $3$, so a pair satisfies one exactly when it satisfies the other. The two recoveries therefore make exactly the same demands on $(x, y)$: they are one system, written at two scales. Isolating $y$ in either produces the very same expression, $y = 2x - 5$, which is why both could have generated the recorded line.

Now separate what was pinned down from what was not. The line fixes the expression that replaced the letter, so it fixes the relationship the isolated equation states, and it fixes which letter was isolated and which equation received it. It says nothing about the size of the numbers each equation happens to be written with, nor about the arrangement of its terms. A recovery is determined as a pair of relationships, not as a pair of written lines.

Turn the hypothesis into an equality between numbers before exchanging anything.

Suppose $(p, q)$ satisfies both equations. The first of them, read at this pair, says

$q = ap + b,$

so for this particular pair the symbol $q$ and the number $ap + b$ are the same number, not merely similar expressions.

The second equation, read at the same pair, says $cp + dq = e$. Replacing a number by an equal number cannot change whether a statement is true, so the statement survives the exchange:

$cp + d(ap + b) = e.$

That is exactly $S$ with $p$ put for $x$. Nothing about $p$ and $q$ was used except that they satisfy the two equations, so every solution of the system contributes a solution of $S$: none has been lost.

Here nothing is solved. A number is handed over and a pair has to be manufactured from it, and the isolation says which second coordinate to attach.

Let $p$ satisfy $S$ and set

$q = ap + b.$

The first equation of the system asks whether $q = ap + b$, and that is how $q$ was built, so it holds with nothing to check.

The second equation asks whether $cp + dq = e$. Replace $q$ by the number it was defined to be:

$cp + dq = cp + d(ap + b) = e,$

the last equality being precisely the statement that $p$ satisfies $S$. So both equations hold at $(p, \ ap + b)$.

Put beside part A this settles the trade. Every solution of the system yields a solution of $S$, and every solution of $S$ yields a solution of the system, with the two constructions undoing each other. The substituted problem is not an approximation of the original; it has exactly the same solutions, which is why solving it is allowed to count as solving the system.

Do the substitution literally. The equation the expression came from is $y = ax + b$, and putting $ax + b$ where $y$ stands turns it into

$ax + b = ax + b.$

This is satisfied by every number, since both sides are the same expression. No value of $x$ can fail it.

Now test the two directions against it. Part A's direction still holds, and trivially so: a solution of the system satisfies an equation that everything satisfies. Part B's direction is where the damage is. It asked that a solution of the derived equation, paired with $ap + b$, satisfy the system, and that now fails, because the derived equation accepts every $p$ while the pairs it produces, $(p, \ ap + b)$, are the whole of the isolated line. The equation $cx + dy = e$ was never used, so nothing it demands has been imposed on anything.

So the lesson's instruction to substitute into the OTHER equation is not etiquette. It is the only version of the move for which both directions of part A and part B can be proved, and the self-substitution loses precisely the half that does the pinning down. The one situation in which the self-substitution does no damage is the one where the second equation was already saying what the first says, and there the whole line really is the answer.

Substitute the pair into each given equation in turn and evaluate fully, rather than stopping at the first one.

In $2x + 3y = 7$:

$2 \cdot \frac{8}{7} + 3 \cdot \frac{11}{7} = \frac{16}{7} + \frac{33}{7} = \frac{49}{7} = 7.$

This one holds. In $3x - y = 5$:

$3 \cdot \frac{8}{7} - \frac{11}{7} = \frac{24}{7} - \frac{11}{7} = \frac{13}{7}.$

That is not $5$, so this one fails. The pair satisfies one of the two given equations and not the other, which is enough to disqualify it: a solution of a system has to satisfy both at once.

Isolate $y$ in $3x - y = 5$ properly and compare. Subtracting $3x$ from both sides gives

$-y = 5 - 3x,$

and the left side is not yet $y$. Multiplying by $-1$ has to reach every term on both sides:

$y = 3x - 5.$

Priya kept the right side as it stood and changed only the left, which is not a move any rule licenses: an operation applied to one side of an equation and not the other does not preserve which pairs satisfy it.

How much damage that does is worth seeing. Her line $y = 5 - 3x$ is the correct isolation of $3x + y = 5$, a different equation from the one she was given, so every line below it is faithful work on a system she was not set. Everything after that line is faithful work on the wrong system, which is why no later check of her arithmetic could have turned anything up. Note also where the fault is NOT: her substitution went into the equation the expression did not come from, exactly as it should.

With the isolation repaired, the route is the ordinary one. Substitute $y = 3x - 5$ into the equation it did not come from:

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

Distribute and collect:

$2x + 9x - 15 = 7, \qquad 11x = 22, \qquad x = 2.$

The isolation returns the other coordinate:

$y = 3(2) - 5 = 1.$

Test $(2, 1)$ in the equations as given:

$3(2) - 1 = 5, \qquad 2(2) + 3(1) = 7.$

Ask what would have had to happen for her test to come out badly.

Her second coordinate was computed from her first by the rule $y = 5 - 3x$, so the test simply runs that computation a second time:

$5 - 3 \cdot \frac{8}{7} = \frac{35 - 24}{7} = \frac{11}{7}.$

The pair was manufactured to satisfy that equation, and putting it back in can only confirm what its construction already guaranteed. A test that cannot fail reports nothing when it passes: it would have passed just as cheerfully if her rearrangement had been wilder still.

The same objection reaches further than she would expect, which is why part A had to evaluate both equations. Her value of $x$ came from solving $2x + 3(5 - 3x) = 7$, so her pair satisfies $2x + 3y = 7$ by construction too, and the calculation in part A confirms it. Two of the three equations on her page were bound to accept the pair.

What is left is the one equation her work never touched again after rearranging it, and that is the only equation of the three capable of refusing her pair. It refuses it, giving $\frac{13}{7}$ where $5$ was wanted. The rule to take away is not merely to check, but to check against the equations exactly as they were given, and against both of them: an equation your own work produced is a witness you coached.