The Substitution Method

Why we need it#

Graphs are lovely for see­ing what's going on, but they let you down when the answer is a frac­tion. The sub­sti­tu­tion method finds the exact solu­tion by pure alge­bra: write one vari­able in terms of the other, sub­sti­tute it in, and you're left with a sin­gle equa­tion in one unknown, which you already know how to solve.

The method in three steps#

1. Take whichever equa­tion is eas­ier and write one vari­able in terms of the other.
2. Sub­sti­tute that expres­sion into the other equa­tion. You now have one equa­tion in one vari­able, solve it.
3. Put the value back into the expres­sion from step 1 to get the sec­ond vari­able.

Then check your answer in both orig­i­nal equa­tions. Check­ing in only one catches noth­ing, because the value came from that very equa­tion.

Worked exam­ple 1#

Solve $x + 3y = 11$ and $2x - y = 1$.

Step 1. The first equa­tion gives $x$ cleanly: $x = 11 - 3y$.

Step 2. Sub­sti­tute into the sec­ond: $2(11 - 3y) - y = 1$, so $22 - 6y - y = 1$, so $-7y = -21$, so $y = 3$.

Step 3. $x = 11 - 3(3) = 2$.

Ver­ify. $2 + 9 = 11$ ✓ and $4 - 3 = 1$ ✓. Solu­tion: $x = 2$, $y = 3$.

Worked exam­ple 2, a frac­tional answer#

Solve $3x + 2y = 5$ and $x - 4y = 4$.

Step 1. From the sec­ond: $x = 4 + 4y$.

Step 2. $3(4 + 4y) + 2y = 5$, so $12 + 12y + 2y = 5$, so $14y = -7$, so $\displaystyle y = -\frac{1}{2}$.

Step 3. $\displaystyle x = 4 + 4\left(-\frac{1}{2}\right) = 4 - 2 = 2$.

Ver­ify. $6 - 1 = 5$ ✓ and $2 + 2 = 4$ ✓.

A graph would have left you squint­ing and guess­ing at that value. Sub­sti­tu­tion hands it to you exactly.

What hap­pens when there is no unique solu­tion#

Here's a nice thing about sub­sti­tu­tion: it never fails silently. When there isn't a sin­gle answer, it hands you a state­ment that is either always true or always false.

Par­al­lel lines. Solve $x + 2y = 3$ and $2x + 4y = 10$. From the first, $x = 3 - 2y$. Then $2(3 - 2y) + 4y = 10$ gives $6 - 4y + 4y = 10$, i.e. $6 = 10$. That is false for every $y$: no solu­tion.

Coin­ci­dent lines. Solve $x + 2y = 3$ and $2x + 4y = 6$. Same sub­sti­tu­tion: $6 - 4y + 4y = 6$, i.e. $6 = 6$. That is true for every $y$: infi­nitely many solu­tions.

So if the vari­able sud­denly van­ishes, don't panic; just look at what is left. A false state­ment means the lines are par­al­lel, and a true one means they coin­cide.

Choos­ing what to sub­sti­tute#

A prac­ti­cal tip: if some vari­able has coef­fi­cient $1$ or $-1$, pick that one. Solv­ing $5x - 2y = 3$ for $x$ gives $\displaystyle x = \frac{3 + 2y}{5}$, which is cor­rect but an open invi­ta­tion to frac­tion errors. If the other equa­tion has $y$ on its own, use that instead and save your­self the trou­ble.

Prac­tice#

Solve by sub­sti­tu­tion and ver­ify.

1. $x + y = 7$, $x - y = 3$.
2. $2x + y = 8$, $x - 3y = -3$.
3. $3x - y = 7$, $2x + 5y = -1$.
4. $x + 2y = 5$, $3x + 6y = 15$.
5. $x - y = 1$, $2x - 2y = 5$.
6. The sum of two num­bers is $32$ and their dif­fer­ence is $8$. Find them.

Answers#