# Isolating quantities on the SAT

### SAT Subscore: Passport to advanced mathematics

## Studying isolating quantities

**On the SAT test isolating quantities topic** is part of passport to advanced mathematics subscore that includes 9 advanced topics (see the full topics list on the top menu).

**Isolating quantities topic is the eighth topic** of passport to advances mathematics subscore. It is recommended to start learning passport to advances mathematics subscore with its first seven topics (the first topic is quadratic equations and quadratic functions).

**Isolating quantities topic is divided into sections** from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.

**Finish studying** heart of algebra subscore topics before you study this topic or any other passport to advances mathematics subscore topic. (Heart of algebra subscore includes basic algebra topics which knowledge is required for understanding passport to advanced mathematics subscore topics).

### Isolating quantities- summary

**Isolating quantities** topic is about isolating variables, so we get a variable equal to an expression containing other variables.

**Isolated variable** is a variable that is written alone on one side of the equation (the second side of the equation contains other variables). For example: In the equation y=mx+b the variable y is isolated.

**Insulating versus solving**: We can isolate any variable of the equation by rearranging the equation according to the same rules we use to solve it. These are the rules for maintaining equality: we can add or subtract the same value from each side of the equation and we can divide or multiply each side of the equation by a same value.

**Note that** the isolated variable will be equal to an expression containing other variables and not a numeric solution like after solving an equation.

**Isolating quantities questions** may ask you to isolate variables in different types of equations, such as linear equations or quadratic equations.

If you are not familiar with solving an equation topic visit the page about linear equations (heart of algebra subscore) before learning this topic.

### Isolating variables in linear equations

You need to follow the rules for maintaining equality, in addition you may need to use distributing and combining like terms. If you need to learn more, see linear equations topic.

Consider the following example:

Slope intercept form of a linear function is given by the equation y=3x+7. What is x in terms of y?

y=3x+7

y-7=3x+7-7

y-7=3x

x=(y-7)/3

Now we can calculate x coordinate given y coordinate area. For example: if y coordinate is 16 then x coordinate is x=(16-7)/3=3.

Checking the answer:

y=3x+7

y=3((y-7)/3)+7

y=y-7+7

y=y

Consider the following example:

5a^{2}-7b=3a^{2}-3b+5

What is b in terms of a?

5a^{2}-7b=3a^{2}-3b+5

-3b+7b=5a^{2}-3a^{2}-5

4b=2a^{2}-5

b=(2a^{2}-5)/4

__Checking the answer:__

5a^{2}-7b=3a^{2}-3b+5

5a^{2}-7(2a^{2}-5)/4=3a^{2}-3(2a^{2}-5)/4+5

Multiply each part of the equation by 4:

20a^{2}-7(2a^{2}-5) =12a^{2}-3(2a^{2}-5)+20

20a^{2}-14a^{2}+35=12a^{2}-6a^{2}+15+20

6a^{2}+35=6a^{2}+35

Consider the following example:

The plant uses function c defined by c(n)=100,000/n+5 to calculate the cost c(n), in dollars, of producing one product. The variable production cost is 5 dollars, the fixed cost of the plant is 100,000 dollars and n is defined as the quantity of the products.

Rewrite n in terms of c.

c=100,000/n+5

c-5=100,000/n+5-5

c-5=100,000/n

n(c-5)=100,000n/n

n(c-5)=100,000

n(c-5)/(c-5)=100,000/(c-5)

n=100,000/(c-5)

Now we can calculate the number of units that were produced given the production cost of a product. For example: if the production cost of a product is 10 dollars, what is the number of products that the plant produced?

n=100,000/(c-5)

n=100,000/(10-5)

n=100,000/5=20,000

Checking the answer:

c=100,000/n+5

c=100,000/(100,000/(c-5))+5

c=100,000*((c-5)/100,000)+5

c=c-5+5

c=c

### Isolating variables in quadratic equations without x term

Since we have only x^{2 }term and there is no x term in the equation, we can isolate x^{2 }

**Step 1:** Isolate x^{2 }with the rules for maintaining equality, in addition you may need to use distributing and combining like terms. If you need to learn more, see linear equations topic.

**Step 2:** Take a square root from each side of the equation. If you need to learn more, see quadratic equations topic.

Consider the following example:

The area of a square is the product of the length l of each side with itself A=l^{2}.

What is l in terms of A?

Step 2: Take a square root of both sides of the equation:

A=l^{2}

√ A=√ l^{2}

√ A=l

l=√ A

Now we can calculate the length of each side of a square given its area. For example: if the area is 49 then the length of its side is √ 49=7.

Checking the answer:

A=l^{2}

A=√ A^{ 2}

A=A

Consider the following example:

5a^{2}-7b=3a^{2}-3b+5

What is a equal to in terms of b?

Step 1: Isolate x^{2 }

5a^{2}-7b=a^{2}-3b+5

5a^{2}-a^{2}=-3b+7b+5

4a^{2}=4b+5

4a^{2}/2=(4b+5)/2

a^{2}=(4b+5)/4

Step 2: Take a square root from each side of the equation

a^{2}=(4b+5)/4

√a^{2}=√(4b+5)/√4

a=√(4b+5)/2

Checking the answer:

5a^{2}-7b=a^{2}-3b+5

5(√(4b+5)/2)^{2}-7b=(√(4b+5)/2)^{2}-3b+5

(5(4b+5))/4 -7b=(4b+5)/4-3b+5

Multiply each part of the equation by 4:

5(4b+5)-28b=4b+5-12b+20

20b+25-28b=4b+5-12b+20

-8b+25=-8b+25

You just finished studying isolating quantities topic, the eighth topic of passport to advanced mathematics subscore!

Continue studying the next passport to advanced mathematics subscore topic- function notation.