# Math Detective: Solving for X (The Two Step Equation)

Two-step equations, as the name implies, can be solved by performing two operations. The key to solving for x is to begin with the end in mind, and end with the beginning in mind.

My “Math Detective” series of essays left off with the case of the one-step equation. Such equations, such as “x + 9 = 22″ and “x^2 = 64″ are one-step equations because they only involve one operation, and therefore require only one step to arrive at a solution.

However, not every equation can be solved with one step. Multi-operational equations will require multiple steps in order to arrive at a solution. To solve these equations, you MUST know the hierarchy of operations, which is commonly called the “order of operations.”

We shall now visit the case of the **two-step equation**. Consider the following…

x^2 + 9 = 90

There are TWO operations present in this equation: exponentiation and addition. Before we discus solving for x, *let us first discuss this equation as if we already kn**o**w **the value of x*. The hierarchy of operations would have us to exponentiate the value of x to the power of two, and then add 9 to the product. Although we do not know the value of x, we know that the result of these operations will be 90.

**Begin with the end in mind, and end with the beginning in mind**

Remember, the detective on a crime scene observes what has happened, and then moves in reverse, tracing back each step of the perpetrator from the end of the crime to the start of the crime. Similarly, the algebraist solves for x by taking the hierarchy of operations and *moving backward, undoing each step*.

Step One: Subtract 9 from both sides |
x^2 + 9 = 90
x^2 = 81 |
Explanation: If we knew what x was and we were computing the result of x2 + 9, then the last step that we would take would be to add 9 to the product of x2. So our first step of solving for x will be to undo the last step by subtracting 9 from both sides. |

Step Two: Take the square root of both sides |
x^2 = 81 x = ±9 |
Explanation: If we knew what x was and we were computing the result of x2 + 9, then our first step would have been to compute x2. So now that we have already done away with the “+ 9″, we are left with the knowledge that x2 = 81. So we will find out the value of x by taking the square root of both sides. This will give us the value of x = ±9 (which means that either 9 or -9 could fit). |

Now that we know that x = 9, let’s look at how the expression x2 + 9 works out once we substitute 9 for x and turn the expression into a 6th grade arithmetic problem…

92 + 9 = 81 + 9 = 90

We squared 9 to get 81, then added 9 to get 90. Our method of finding the value of x undone every step that would have been carried out. Begin with the end in mind, and end with the beginning in mind.

Let’s do another example…

2x – 5 = 5

Step One: Add 5 to both sides |
2x – 5 = 5
2x = 10 |
Explanation: If we knew the value of x in the expression 2x - 5, then we would first multiply x by 2, and then subtract five from the product. So well will begin solving for x by undoing the last step and add 5 to both sides. |

Step Two: Divide both sides by 5 |
x = 5 |
Step one left us with 2x = 10. We now know that 2 times whatever x is will equal ten. By undoing the last step of multiplication by to with dividing by two, we are left with x = 5. |

Finally, we will examine one last case…

√(x – 3) = 2

Step One: Raise both sides to the power of two |
√(x - 3) = 2 (√(x - 3))^2 = 2^2 x – 3 = 4 |
Explanation: If we were computing √(x - 3) and we knew the value of x, then the first thing that we would do would be to evaluate what is inside the parenthesis, followed by computing the square root. So we begin with undoing the final step of taking the square root of (x – 3) by raising both sides to the power of two. |

Step Two: Add three to both sides |
x – 3 = 4
x = 7 |
We now know that the radicand x – 3 is equal to 7. We will now undo the – 3 by adding 3 to both sides to get x = 7. |

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