## Homework: Solve This Equation 4 Ways

Posted by mark schwartz on October 2, 2016

__Introduction__

I once had a class of pre-service and in-service teachers and one of our discussions led us to explore using one’s imagination to work through a math problem, rather than relying on the standard, traditional algorithm. I asked if all of them had taken Algebra and were comfortable solving equations. As it turns out they, collectively, were quite adept at it. So, the following story shows not only how adept they were but also how imaginative they were.

__The Story__

First let me note that they worked in groups, so the following homework assignment was started in class and I urged them to swap email addresses so they could work on it at home, which they did. Here’s the homework: you are to solve 4 ─ 2(x ─ 1) = 8 + 6(x ─ 1) 4 different ways and annotate the steps in your solution. We reviewed the traditional algorithm in class, so it didn’t count, but I did include it in the following list of solutions.

At our next class, I asked them to write their alternative solutions on the board and although there were many similar solutions, listed below are some of the unique ones. As we discussed these, the question was asked “when teaching math, should we ever have students do an exercise like this?”

Some felt we should; some felt otherwise. The critical difference was whether or not their school’s curriculum would allow for this. They liked the idea of using imagination and had fun doing the exercise but they collectively concluded that students in elementary and secondary school should stick to the traditional. I asked them that if a student, unprompted and independently, worked an equation – or solved any other mathematical expression – in an unorthodox way, what would they do?

It was a good discussion with no clear closure about what to do with the unorthodox student. Given the current press in Common Core Math, where students are to express their reasoning, it seems that accepting unorthodox solutions might be reasonable, but on the other hand a student may have an “aha” moment and can’t clearly articulated how the solution was found. It’s an interesting challenge for today’s teachers.

Here are some of the results of their work, annotated. Follow what they did because clearly imagination was in play.

__Traditional:__

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

4 ─ 2x + 2 = 8 + 6x ─ 6

6 ─ 2x = 2 + 6x

4 = 8x

1/2 = x

__1st Alternative (only a slight difference, but a difference):__

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

2 ─ (x ─ 1) = 4 + 3(x ─ 1) divide all terms by 2

2 ─ x + 1 = 4 + 3x ─ 3 do the indicated multiplications on both sides

3 ─ x = 1 + 3x combine like terms on both sides

2 ─ x = 3x subtract 1 from both sides

2 = 4x add x to both sides

1/2 = x divide both sides by 4 and simplify answer

__2nd Alternative:__

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

0 = 4+8(x─1) subtract 4 ─ 2(x ─ 1) from both sides

0 = 4 + 8x ─ 8 do indicated multiplication

0 = ─4 + 8x combine like terms

4 = 8x add 4 to both sides

1/2 = x divide both sides by 8

__3rd Alternative:__

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

4 = 8 + 8(x ─ 1) add 2(x ─ 1) to both sides

─4 = 8(x ─ 1) subtract 8 from both sides

─ 1/2 = x ─ 1 divide both sides by 8

1/2 = x add 1 to both sides

__4th Alternative:__

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

4/(x ̶ 1) ̶ 2 = 8/(x ̶ 1) + 6 divide every term by (x ─ 1)

─ 2 = 4/(x ̶ 1) + 6 subtract 4/(x ̶ 1) from both sides

─ 8 = 4/(x ̶ 1) subtract 6 from both sides)

─ 8(x ─ 1) = 4 multiply both sides by (x ─ 1)

─ 8x + 8 = 4 do the indicated multiplication on the left side

─ 8x = ─ 4 subtract 8 from both sides

x = 1/2 divide both sides by ─ 8

__5 ^{th} Alternative__

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

1/2 ─ (1/4)(x ─ 1) = 1 + (3/4)(x ─ 1) divide every term by 8

1/2 = 1+ x ─ 1 add (1/4)(x ─ 1) to both sides

1/2 = x combine terms on the right side

__6th Alternative:__

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

4 ─ 2x = 8 + 6x let x = x ─ 1

─ 4 = 8x subtract 8 from both sides; add 2x to both sides

─ 1/2 = x divide both sides by 8 and simplify

─ 1/2 = x ─ 1 let x ─ 1 = x (“reverse” first operation)

1/2 = x add 1 to both sides

__7th Alternative:__

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

4(x + 1) ─ 2(x ─ 1) (x + 1) = 8(x + 1) + 6(x ─ 1)(x + 1) multiply every term by (x + 1)

4x + 4 ─ 2x^{2} + 2 = 8x + 8 + 6x^{2} ─ 6 do all indicated multiplications

─ 2x^{2} + 4x + 6 = 6x^{2} +8x +2 combine like terms on both sides

─ x^{2} + 2x + 3= 3x^{2} +4x + 1 divide all terms by 2

(─x + 3)(x + 1) = (3x + 1)(x + 1) factor both sides

─x + 3 = 3x + 1 divide both sides by (x + 1)

3 = 4x + 1 add x to both sides

2 = 4x subtract 1 from both sides

1/2 = x divide both sides by 4 and simplify

__8th Alternative:__

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

16 ─ 16(x ─ 1) + 4(x ─ 1)^{2} = 64 + 96(x ─ 1) + 36(x ─ 1)^{2} square both sides

16 ─ 16x + 16 + 4x^{2} ─ 8x + 4 = 64 + 96x ─ 96 + 36x^{2 }─ 72x +36 do all indicated operations

4x^{2} ─ 24x + 36 = 36x^{2} + 24x + 4 combine like terms

x^{2} ─ 6x + 9 = 9x^{2} + 6x + 1 divide every term by 4

0 = 8x^{2} + 12x ─ 8 subtract x^{2} ─ 6x + 9 from both sides

0 = 2x^{2} + 3x ─ 2 divide every term by 4

0 = (2x ─ 1)(x + 2) factor 2x^{2} + 3x ─ 2

1/2 = x setting both factors equal to 0 gives…

(x = ─ 2 is an extraneous root

Introduced when both sides were

Squared

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