4. Solving Equations With Variables on Both Sides
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| A | B | C | |
|---|---|---|---|
| 1 | x | 7(x+1) | 3(x-1) |
| 2 | -5 | -28 | -18 |
| 3 | -3 | -14 | -12 |
| 4 | -1 | 0 | -6 |
| 5 | 1 | 14 | 0 |
| 6 | 3 | 28 | 6 |
A value is a solution for an equation if substituting it into the equation makes a true statement. In this case, we are evaluating the expressions 7(x+1) and 3(x-1) for certain values of x. If one of them was the solution for the equation 7(x+1) = 3(x-1), then both expressions would produce the same number for that specific x-value. We can see that this does not happen. Therefore, the solution isn't shown.
| A | B | C | D | |
|---|---|---|---|---|
| 1 | x | 7(x+1) | 3(x-1) | | 7(x+1) - 3(x-1)| |
| 2 | -5 | -28 | -18 | 10 |
| 3 | -3 | -14 | -12 | 2 |
| 4 | -1 | 0 | -6 | 6 |
| 5 | 1 | 14 | 0 | 14 |
| 6 | 3 | 28 | 6 | 22 |
As we can see, the value of 7(x+1) was getting closer to the value of 3(x-1) until x reached the value -1. After that, their values started to get further away from one another. This means that the solution must be between -3 and -1.
| A | B | C | D | |
|---|---|---|---|---|
| 1 | x | 7(x+1) | 3(x-1) | 7(x+1) < 3(x-1) |
| 2 | -5 | -28 | -18 | ✓ |
| 3 | -3 | -14 | -12 | ✓ |
| 4 | -1 | 0 | -6 | * |
| 5 | 1 | 14 | 0 | * |
| 6 | 3 | 28 | 6 | * |
We can see that this is only true for the values x=-5 and x= -3.