# Solving Formulas for Specific Variables

Hello! Welcome to this video on solving equations for specific variables. This is an algebraic topic that has a lot of practical applications in the science fields.

For example, let’s say we have Newton’s second law of motion:

$$F=ma$$

and instead of being given the mass and acceleration to find the force, we are given the force and the acceleration and are asked to find the mass. All we have to do is rearrange this formula to solve for the variable $$m$$. So, since $$m$$ is being multiplied by $$a$$, we need to divide both sides by $$a$$ to undo this multiplication.

$$\frac{F}{a}=\frac{ma}{a}$$

So on the left side we’ll have $$\frac{F}{a}$$, and on the right side our $$a$$‘s cancel out, and we’re left with $$m$$.

$$\frac{F}{a}=m$$

So this is our equation. Now we can easily plug in our values for force and acceleration and find our mass.

What if we’re given the equation:

$$e=mc^{2}$$

and we’re asked to solve for $$c$$? Well, there’s multiplication between $$m$$ and $$c^{2}$$, and a square that we need to undo. When simplifying equations, we use the order of operations (PEMDAS). To rearrange equations, we need to do this backwards, so we need to first undo multiplication and then undo the square (or the exponent). So, we said we need to undo multiplication first, so we’re multiplying $$m$$ by $$c^{2}$$, so we need to do the opposite of multiplication, which is division, and divide both sides by $$m$$.

$$\frac{e}{m}=\frac{mc^{2}}{m}$$

So this will give us, $$\frac{e}{m}=c^{2}$$, because our $$m$$‘s cancel out.

Now we need to undo the square by square rooting both sides.

$$\sqrt{\frac{e}{m}}=\sqrt{c^{2}}$$

So if we take the square root of both sides, we can leave this left side as $$\sqrt{\frac{e}{m}}$$, and if you have $$\sqrt{c^{2}}$$, it cancels out, which is what we were wanting. And we’re left with $$c$$.

$$\sqrt{\frac{e}{m}}=c$$

Let’s try one more example. This one doesn’t have a science application like the other ones do, but it’s something you might see in your algebra class.

$$3xy-4=2z$$

And I want you to solve for the variable $$y$$.

So, remember, we need to do the reverse order of operations and we need to get rid of everything on this side, except for the $$y$$, because that’s what we’re solving for. So we’re going to start by adding 4 to both sides.

$$3xy-4+4=2z+4$$

These terms cancel out on the left and we’re left with:

$$3xy=2z+4$$

Now, since both 3 and $$x$$ are being multiplied by $$y$$, we can simply divide both sides of our equation by $$3x$$. We can do the division of the 3 and the division of the $$x$$ at the exact same time, and that’s what this will look like. When we do this though, make sure you divide the entire right side by $$3x$$. If you only do the $$2z$$ part or only the 4 part, you’ll get the wrong answer, so make sure you do the entire thing by $$3x$$.

$$\frac{3xy}{3x}=\frac{2z+4}{3x}$$

So, over here our $$3x$$ terms cancel out and we’re left with $$y$$.

$$y=\frac{2z+4}{3x}$$

I hope this was helpful. Thanks for watching and happy studying!

## Practice Questions

Question #1:

The equation of a line in slope-intercept form is:

$$y=mx+b$$

where $$m$$ is the slope of the line, $$b$$ is the $$y$$-intercept of the line, and $$x$$ and $$y$$ are the coordinates of a point $$(x,y)$$ on the line. Solve the equation for $$m$$.

$$m=\frac{y}{b}-x$$
$$m=\frac{y-x}{b}$$
$$m=\frac{y-b}{x}$$
$$m=\frac{y}{x}-b$$

We can rearrange the equation using algebraic methods such as adding, subtracting, multiplying, or dividing terms to both sides of the equation to solve for the specified variable.

To isolate the variable $$m$$, start by subtracting $$b$$ from both sides.

$$y-b=mx+b-b$$
$$y-b=mx$$

Then, divide by $$x$$ on both sides.

$$\frac{y-b}{x}=\frac{mx}{x}$$

$$\frac{y-b}{x}=m$$

Therefore, $$m=\frac{y-b}{x}$$.

Question #2:

The equation for the area, $$A$$, of a trapezoid is given below in terms of its height, $$h$$, and two bases, $$b_1$$ and $$b_2$$.

$$A=\frac{1}{2}(b_1+b_2)h$$

Solve the equation for $$b_1$$.

$$b_1=\frac{2A}{h}-b_2$$
$$b_1=\frac{2A-b_2}{h}$$
$$b_1=\frac{Ah}{2}-b_2$$
$$b_1=\frac{A}{2b_2}-h$$

We can rearrange the equation using algebraic methods such as adding, subtracting, multiplying, or dividing terms to both sides of its equation to solve for the specified variable.

To solve the equation for $$b_1$$, start by multiplying both sides by 2.

$$\mathbf{2}\cdot A=\mathbf{2 \cdot}\frac{1}{2}\left(b_1+b_2\right)h$$

$$2A=(b_1+b_2)h$$

Next, divide both sides by $$h$$.

$$\frac{2A}{\mathbf{h}}=\frac{(b_1+b_2)h}{\mathbf{h}}$$

$$\frac{2A}{h}=b_1+b_2$$

Finally, subtract $$b_2$$ from both sides.

$$\frac{2A}{h}\mathbf{-b_2}=(b_1+b_2)\mathbf{-b_2}$$

$$\frac{2A}{h}-b_2=b_1$$

Therefore, $$b_1=\frac{2A}{h}-b_2$$.

Question #3:

The equation for the volume, $$V$$, of a right circular cone is given below in terms of its radius, $$r$$, and its height, $$h$$.

$$V=\frac{1}{3}\pi r^2h$$

Solve the equation for the cone’s radius.

$$r=\frac{\sqrt{\pi V}}{3h}$$
$$r=\frac{\sqrt{3V}}{\pi h}$$
$$r=\sqrt{\frac{\pi V}{3h}}$$
$$r=\sqrt{\frac{3V}{\pi h}}$$

Solve for the radius, $$r$$, by first multiplying both sides by 3

$$\mathbf{3}\cdot V=\mathbf{3}\cdot\frac{1}{3}\pi r^2h$$

$$3V=\pi r^2h$$

Then, divide both sides by $$\pi h$$.

$$\frac{3V}{\mathbf{\pi h}}=\frac{\pi r^2h}{\mathbf{\pi h}}$$

$$\frac{3V}{\mathbf{\pi h}}=r^2$$

Finally, take the square root of both sides.

$$\sqrt{\frac{3V}{\pi h}}=\sqrt{r^2}$$

$$\sqrt{\frac{3V}{\pi h}}=r$$

Therefore, $$r=\sqrt{\frac{3V}{\pi h}}$$.

Question #4:

The temperature of an outdoor thermometer reads $$–4°$$ (negative 4 degrees Celsius). The equation to find the temperature in degrees Celsius, $$C$$, given the temperature in degrees Fahrenheit, $$F$$, is:

$$C=\frac{5}{9}(F-32)$$

You want to know the temperature outside in degrees Fahrenheit. Which of the following equations can be used to directly find the temperature in degrees Fahrenheit?

$$F=\frac{9}{5}(C+32)$$
$$F=\frac{9}{5}C+32$$
$$F=\frac{5}{9}(\frac{C}{F-2})$$
$$F=\frac{5}{9}C+32$$

To find the temperature in degrees Fahrenheit in terms of the temperature in degrees Celsius, we need to solve for $$F$$. Start by multiplying both sides by $$\frac{9}{5}$$.

$$\mathbf{\frac{9}{5}\cdot} C=\mathbf{\frac{9}{5}\cdot}\frac{5}{9}(F-32)$$

$$\frac{9}{5}C=F-32$$

From here, add 32 to both sides.

$$\frac{9}{5}C\mathbf{+32}=F-32\mathbf{+32}$$

$$\frac{9}{5}C+32=F$$

Therefore, the formula for finding the temperature in degrees Fahrenheit when given the temperature in degrees Celsius is $$F=\frac{9}{5}C+32$$.

Question #5:

The cost, $$C$$, in dollars, for a company to produce x textbooks is given by the equation:

$$C=3(x+10)-12$$

Which of the following equations can be used to directly find the number of textbooks produced at a given cost?

$$x=\frac{C}{x+10}+12$$
$$x=\frac{C}{3}+4$$
$$x=\frac{C+12}{3}-10$$
$$x=\frac{C}{3}+2$$

To find the number of textbooks in terms of their cost to produce, we need to solve for $$x$$. Start by adding 12 to both sides of the equation.

$$C\mathbf{+12}=3(x+10)-12\mathbf{+12}$$
$$C+12=3(x+10)$$

From here, divide both sides by 3.

$$\frac{C+12}{\mathbf{3}}=\frac{3(x+10)}{\mathbf{3}}$$

$$\frac{C+12}{3}=x+10$$

Then, subtract 10 from both sides.

$$\frac{C+12}{3}-10=x+10-10$$

$$\frac{C+12}{3}-10=x$$

Therefore, the number of textbooks in terms of their cost to produce is $$x=\frac{C+12}{3}-10$$.