Apa itu persamaan dengan banyak variabel?

Literal Equations

Like any equation, you can often solve literal equations for one variable, but another approach is to guess-and-check, or by trial and error.

But before we pull out the calculator and do this, we need to talk about what literal equations are.

Literal equation are equations (meaning they have an equals sign) that contain more than one distinct variable. They can be linear equations, quadratic equations, cubic equations, etc. But be careful — multiple appearances of the same variables in the equation don't count. For instance, $$x^2 + y^2 = 1$$ is a literal equation because $x$ and $y$ both appear in the equation. On the other hand $$x^2 + x = 1$$ is NOT a literal equation because $x$ is the only variable that appears.

Here's why we connect these types of equations to finding solutions to these equations with trial-and-error. Literal equations have more than one variable (and often at least one variable on each side of the equals sign), and many equations with only one variable can be solved without trial-and-error. So we are trying to focus on the equations that can't easily be solved without trial-and-error.

Let's go back to our first example of a literal equation, $$x^2 + y^2 = 1.$$

We want to try and find some solutions to this equation. One way to do this is to guess-and-check by substituting certain values of $x$ and $y$ to see if the equation works. A good place to get started is by letting some or all of the variables equal $0,$ $1,$ or $-1,$ because the algebra that follows is usually easier.

Perhaps $x = 0.$ If so the equation is $$y^2 = 1.$$

We could take the square root of this to see that if $x = 0,$ then $y = 1$ OR $y = -1.$ Thus two solutions are $(0, 1)$ and $(0, -1).$ Let's try finding more.

Maybe $x = 1,$ and then $$1 + y^2 = 1,$$ and this can be simplified to see $y^2 = 0,$ or $y = 0,$ meaning $(1, 0)$ is a solution. If we let $x = -1,$ then the same steps show us that $(-1, 0)$ is ALSO a solution.

Wow, we found four solutions, just like that. So why don't we always use trial and error?

Trial-and-error lets us check whether given values of the variables are solutions to the equation, but it tells us nothing about whether there are other solutions. We have to just keep using trial-and-error, sometimes forever, without knowing if we've found ALL solutions. This can sometimes be a major downside, but sometimes we don't care how many solutions there are if we can only find one.

In the problem we just did, we set you loose to find solutions to a literal equation. Some questions will be a bit more guided when it comes to finding solutions. One common type of multiple choice question is to have a literal equation with a bunch of answer choices, asking which answer choices are solutions to the equation. In this case you just have to plug in the answer choices given to you and check each of them.

Like I said before, it comes down to plugging in $x$ and $y$ to see if the equation works. Do this for each answer choice.

First answer choice: plug in $x = 2$ and $y = 1:$ $$2^2 + 2 + 1 = 4 + 2 + 1 = 7 \neq 5.$$ So the first answer choice is not a solution.

Second answer choice: plug in $x = 2$ and $y = -1:$ $$2^2 + 2 - 1 = 4 + 2 - 1 = 5.$$ So the second answer choice is a solution.

Third answer choice: plug in $x = -2$ and $y = 1:$ $$(-2)^2 + (-2) + 1 = 4 - 2 + 1 = 3 \neq 5.$$ So the third answer choice is not a solution.

Fourth answer choice: plug in $x = -2$ and $y = 3:$ $$(-2)^2 + (-2) + 3 = 4 - 2 + 3 = 5.$$ So the fourth answer choice is a solution.

(Video) Formulas

by mathman1024

This by mathman1024 explains formulas which are also called literal equations.

$\large{\text{Summary}}$

A formula is an equation that describes a known relationship among several variables. You can solve a formula by performing algebraic operations on it and plugging in values that you already know! (Actually, "solving a formula" isn't really what you're doing — you're really just solving for a variable IN the formula.)

Examples of formulas:

• Formula for the circumference of a circle: $C=2\pi r$
• Formula for the area of a triangle: $A=\frac12bh$
• Ideal Gas Law: $I=nRT$
• Simple Interest Formula: $I=PRT$

One formula used often relates distance, rate, and time: $$\text{distance}=\text{rate}\times\text{time} \text{ or } d=rt$$

Say you need to drive a distance of $800\text{ miles}$ and your average rate of speed is $60\text{ miles per hour}$, can you find out how long it will take to get there? Yes you can! You just plug in the values that are known and solve for the variable that is unknown. $$\begin{align} d&=rt\\ ({\color{#8925AE}{800}})&=({\color{#8925AE}{60}})t\\ \frac{800}{{\color{#8925AE}{60}}}&=\frac{60t}{{\color{#8925AE}{60}}}\\ \frac{40}3&=t \end{align}$$ (For a reminder of why we divided both sides by $60,$ here is a refresher on solving equations using division.) So you know it'll take $\frac{40}{3}\text{ hours}=13\frac13\text{ hours}$.

The next example he gives deals with perimeter. The perimeter of a rectangle is: $P=2l+2w$. So a farmer approaches you asking to help him build a fence. He has $220\text{ feet}$ of fence and he knows he wants the length to be $72\text{ feet}$. Can you isolate the variable $w$ to figure out what the width should be so that you use all the fence? Let's find out! $$\begin{align} P&=2l+2w\\ ({\color{#8925AE}{220}})&=2({\color{#8925AE}{72}})+2w\\ 220&=144+2w\\ {\color{#8925AE}{-144}}&={\color{#8925AE}{-144}}\\ \hline 76&=2w\\ \frac{76}{{\color{#8925AE}{2}}}&=\frac{2w}{{\color{#8925AE}{2}}}\\ 38&=w\\ w&=38 \end{align}$$

Well now the farmer comes back. He says he messed up and doesn't want the length to be $72\text{ feet}$ and instead wants it to be $45\text{ feet}$. Well now we need to recalculate this. $$\begin{align} P&=2l+2w\\ ({\color{#8925AE}{220}})&=2({\color{#8925AE}{45}})+2w\\ 220&=90+2w\\ {\color{#8925AE}{-90}}&={\color{#8925AE}{-90}}\\ \hline 130&=2w\\ \frac{130}{{\color{#8925AE}{2}}}&=\frac{2w}{{\color{#8925AE}{2}}}\\ 65&=w\\ w&=65 \end{align}$$

Uh-oh! You see the farmer coming back again. He's probably going to change the length again. You're on to him though. The amount of fence is the same just the length changes. So we can solve the equation for the width. $$\begin{align} 220&=2l+2w\\ {\color{#8925AE}{-2l}}&={\color{#8925AE}{-2l}}\\ 220-2l&=2w\\ 2w&=220-2l\\ \frac{2w}{{\color{#8925AE}{2}}}&=\frac{220-2l}{{\color{#8925AE}{2}}}\\ \frac{2w}{2}&=\frac{220}{2}-\frac{2l}{2}\\ w&=110-l \end{align}$$ As you see we solved this equation for the width. Now when the farmer comes back and gives us a new length, we can just easily throw it into our new equation and out pops the width. If we check our previous lengths with our new formula we get the same answers. $$\begin{align} w&=110-l\\ w&=110-({\color{#8925AE}{72}})\\ w&=38 \end{align}$$ $$\begin{align} w&=110-l\\ w&=110-({\color{#8925AE}{45}})\\ w&=65 \end{align}$$

Literal Equations

Literal equations are equations with more than one variable. Most of the time, you'll be asked to solve for one of the variables using information about another.

So, if you see an equation with $x, y,$ and $z$ variables in it, maybe even on each side of the equation, that's a literal equation. If you see an equation with $a, b, c, d, e, f,$ and $g$ variables in it (may the Math Force be with you), that's still a literal equation.

Examples:

$$3x+5y=6\\ 2k + 3w = r+f$$

Here's an example of a literal equation attached to a short word problem:

*You are headed to the store to buy sandwiches and sodas for a party. Each sandwich costs $9\text{ dollars}$ and each soda costs $3\text{ dollars}$. You have a total budget of $90\text{ dollars}$. How many sodas can you buy if you buy $6\text{ sandwiches}$? Write an equation to model this information. *

We know we have to balance how many sodas we buy and how many sandwiches we buy, so we'll need something that adds $x$ (sandwiches) and $y$ (sodas). Since we're limited based on how much money we have, we can set that equal to $90$. Then, since we know how much sandwiches cost ($\$9$) and how much sodas cost ($\$3$) we start by multiplying $\$9/\text{sandwich}\cdot x\text{ sandwiches}$ and $\$3/\text{soda}\cdot y\text{ sodas}$: $$9x + 3y = 90$$ (We could also think of each sandwich we buy as subtracting from the total budget and get $90 - 9x = 3y$. Both equations are equivalent, so they both work!)

This equation is literal because there is more than one variable involved.

Once you've got this equation, you can solve problems like the one asked: "How many sodas can you buy if you buy $6\text{ sandwiches}$?"