__Transforming Equations__

The makers of the SAT expect you to know basic algebra rules on how to transform and solve equations. You use these skills on about a third of the math problems on the test, making knowing them all critically important. In general, you will see three types of straight-up algebra problems:

- Questions that ask you to isolate a variable
- Questions that ask you to solve for a single variable
- Questions that ask you to manipulate polynomials (equations where the variable has a power greater than 1, like \(x^2\) or \(z^4\)

We’ll address these in order in this section.

__Isolating a Variable__

For instance, you may see:

A communications engineer uses the formula* p = dhw* to find the power*, p *when a transmitter is* d *miles away on top of a tower of* h *height, transmitting at* w *watts. Which of the following correctly expresses* d *in terms of* h,w, *and* p?*

A) *d = phw*

B) \(d=\frac{pw}h\)

C) \(d=\frac{wh}p\)

D) \(d=\frac p{wh}\)

**Strategy:**

**Whenever you see the verbiage “***correctly expresses __ in terms of*”, you should translate this to “*solve for ___*”. In this case, we just need to solve for, or isolate,*d*. We would divide by*h*and*w*to get*d*by itself.**You should always be writing out the steps one by one, rather than doing these problems in your head. These are very easy problems for the most part, but students often slip up when they don’t write down the steps.**

The most fundamental rule you must apply: **always do the exact same thing to both sides of the equation!**

Another example:

*If *\(t^{-3}=x\)* where t > 0, what is t in terms of x?*

A) \(x\sqrt x\)

B) \(\sqrt[3]{\frac1x}\)

C) \(3x\)

D) \(x^{1/3}\)

The solution to this problem is a little trickier, but the way to go about it is the same – we need to get *t* isolated with no exponent. If you don’t remember all your exponent rules right now, don’t worry, those will be covered in a later unit. First, we need to make our exponent positive by moving it to the other side of the divisor to get:

$$\frac1{t^3}\;=x$$

Then we multiply both sides by \(t^3\), to get:

$$1=t^3x$$

we can then divide by x to get:

$$\frac1x=t^3$$

Taking the cube root of both sides gets us to answer B).

**Remember that you must always get the variable by itself, with no exponent, radical, coefficient, or divisor.**

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