Substitution

Worked example

7.2.w1 Example 1. Solve y = 2x and x + y = 9. Here y is already alone. The first equation tells you the box holds 2x. So in the second equation, wherever you see y, pour in 2x: $$\begin{aligned} &x + 2x = 9 \\ &\Rightarrow\; 3x = 9 \\ &\Rightarrow\; x = 3. \end{aligned}$$ Why this works: x plus 2x is three x's, so 3x = 9, and dividing both sides by 3 gives x = 3. But you're not done. You have x, not the pair. Go back and find y: since y = 2x, y = 2(3) = 6. So the solution is (3, 6). Check both: is 6 = 2(3)? Yes. Is 3 + 6 = 9? Yes. Solution: (3, 6).

7.2.w2 Example 2, isolate first. Solve x = y + 1 and 2x + y = 8. This time x is the one already alone: x = y + 1. Substitute y + 1 in place of x in the second equation: $$\begin{aligned} &2(y + 1) + y = 8 \\ &\Rightarrow\; 2y + 2 + y = 8 \\ &\Rightarrow\; 3y + 2 = 8 \\ &\Rightarrow\; 3y = 6 \\ &\Rightarrow\; y = 2. \end{aligned}$$ Step by step: the 2 outside the parentheses has to reach everything inside, so 2(y + 1) becomes 2y + 2. That's the distributing move from Unit 6, handing a copy to each term. Then 2y + y is 3y, so 3y + 2 = 8; subtract 2 from both sides to get 3y = 6; divide by 3 to get y = 2. Now find the partner: x = y + 1 = 2 + 1 = 3. Solution (3, 2). Check both: is 3 = 2 + 1? Yes. Is 2(3) + 2 = 8? Yes, since 6 + 2 = 8. Solution: (3, 2).

That distributing step is the one to slow down on. It's tempting to write 2(y + 1) as 2y + 1, sending the 2 to only the first thing inside. But the 2 has to greet everyone in the parentheses, so it's 2y + 2. A quick way to catch it: count the terms inside, and make sure the outside number reached every one.

7.2.w3 Example 3, watch the distribution. Solve y = x − 2 and x + y = 10. y is alone, so substitute x − 2 for y in the second equation: $$\begin{aligned} &x + (x - 2) = 10 \\ &\Rightarrow\; 2x - 2 = 10 \\ &\Rightarrow\; 2x = 12 \\ &\Rightarrow\; x = 6. \end{aligned}$$ The x plus another x makes 2x, and the −2 comes along, so 2x − 2 = 10; add 2 to both sides for 2x = 12, then divide by 2 to get x = 6. Partner: y = x − 2 = 6 − 2 = 4. Solution (6, 4). Check both: is 4 = 6 − 2? Yes. Is 6 + 4 = 10? Yes. Solution: (6, 4).

7.2.w4 Example 4, isolate when nothing is alone yet. Solve x + y = 7 and 2x + y = 11 by substitution. Neither variable is alone here, so make one alone first. The first equation is the easy one to rearrange: subtract x from both sides to get y = 7 − x. Now substitute 7 − x for y in the second equation: $$\begin{aligned} &2x + (7 - x) = 11 \\ &\Rightarrow\; x + 7 = 11 \\ &\Rightarrow\; x = 4. \end{aligned}$$ Here 2x − x is just x, so x + 7 = 11, and subtracting 7 gives x = 4. Partner: y = 7 − x = 7 − 4 = 3. Solution (4, 3). Check both: is 4 + 3 = 7? Yes. Is 2(4) + 3 = 11? Yes, since 8 + 3 = 11. Solution: (4, 3). (This same system has a matching pair of y-terms, which makes it a natural fit for the elimination method coming in 7.3. You'll solve it that way too.)

Which equation you substitute into matters too. You isolated y from the first equation, so you pour the result into the second one. If you pour it back into the first, the one you just got it from, everything collapses to something like 0 = 0, which is true but tells you nothing, because you've only said "the first equation equals itself." Always substitute into the other equation.

Here's a clean one to get the method moving before the practice mixes things up. Solve y = 5 and x + y = 8. The first equation hands you y outright, so substitute 5 for y in the second: x + 5 = 8, which gives x = 3. Partner's already known, y = 5. Solution (3, 5). Check both: 5 = 5, and 3 + 5 = 8. When one equation just states a value like that, substitution is almost no work at all.