用代入法求解

(Video) Solving Systems of Equations with the Substitution Method

by NancyPi

This video by NancyPi explains systems of equations and substitution.

$\large{\text{Summary}}$

Say we want to solve a system of equations, like $$\begin{align} 2x+y&=11\\ 7x-4y&=-14\\ \end{align}$$

She boils it down to three simple steps:

1) Solve for $x$ or $y$ in either equation. $$\begin{align} 2x+y&=11\\ y&=11-2x \\ \end{align}$$

2) Substitute this solution into the other equation and solve for the second variable. $$\begin{align} 7x-4y&=-14\\ 7x-4(11-2x)&=-14\\ x&=\boxed{2}, \text{ after some algebra} \end{align}$$

3) Use that second solution to solve for the first variable. $$\begin{align} 2x+y&=11\\ 2(2)+y &= 11\\ 4+y &=11\\ y&=\boxed{7} \end{align}$$

For a really in-depth explanation of each of the steps, go ahead and watch the video!

Now, use what you've learned to solve the following system of equations:

(Notice that there isn't just a variable hanging out on its own. Step 1 will take a couple moves!)

Let's work through this problem to solve for $x$ and $y.$ As a reminder, the system of equations we are solving is $$\begin{align} 27 &= 8x + 3y\\ 6y &= 4x + 39 \end{align}$$

Step One: Solve for $x$ or $y$ in either equation

For example, let's solve for $y$ in the first equation. $$\begin{align} 27 &= 8x + 3y\\ \\ 27-8x &= 3y\\ \\ \frac{27-8x}{3}&=y \end{align}$$

Step Two: Substitute this solution into the other equation and solve for the second variable

So, we will substitute $y=\frac{27-8x}{3}$ into our second equation to solve for $x.$

$$\begin{align} 6y &= 4x + 39 \\ 6\bigg(\frac{27-8x}{3}\bigg) &= 4x+39\\ \\ 2\bigg(\frac{27-8x}{3}\bigg) &=4x+39\\ \\ 54-16x &= 4x+39\\ \\ 15-16x &= 4x\\ \\ 15&= 20x\\ \\ x&= \boxed{\frac{3}{4}} \end{align}$$

Step Three: Use that second solution to solve for the first variable

We can substitute $x=\frac{3}{4}$ back into either of the original equations to solve completely for $y.$ Let's substitute it into the first equation. $$\begin{align} 27 &= 8x + 3y\\ \\ 27 &= 8\bigg(\frac{3}{4}\bigg)+3y\\ \\ 27 &= 2(3)+3y\\ \\ 27 &= 6+3y\\ \\ 21 &= 3y\\ \\ y&= \boxed{7} \end{align}$$

There we have it! Our solutions to the system are $x=\frac{3}{4}$ and $y=7.$ Remember to plug these solutions back into the original equations to make sure that they work. This step is very important, because sometimes solutions are not actually valid.

A system of equations is "two or more equations that have the same variables with the same values."

The relationship between equations in a system is pretty cool. You can substitute a value from one equation to solve the second equation. This is called substitution.

The value of the variables will be the same in both equations. This means that $x=5$ in the first equation, $x$ is $5$ in the second equation. Let's look at an example.

$$\begin{align} 3x-y&=7 \\\ 2x+3y&=1 \end{align}$$

The first step to solve a system of equations is to pick an equation and solve for one of the variables. Which equation you pick and which variable you solve for are up to you.

Let's use the first equation because it looks easy to solve for $y$. The easiest approach usually to pick the equation that has a lone variable. That is, a variable that has no coefficient.

$$\begin{align} 3x - y&=7 \\\ -y&=7-3x \\\ y &= 3x - 7 \end{align}$$

Plug the newly found value into the second question. In this case, we substitute $y$ with $3x-7$ in the second equation.

$$\begin{align} 2x+3y&=1 \\\ 2x + 3 (3x - 7) &= 1 \\\ 2x + 9x - 21 &= 1 \\\ 11x &= 22 \\\ x &= 2 \end{align}$$

Nice! We've figured out the value of $x$. Now, let's substitute this value for the $x$ in either of our equations. We'll pick the first one, $3x-y=7$

$$\begin{align} 3x - y &= 7 \\\ 3(2) - y &= 7 \\\ 6-y &= 7\\ -y &= 1 \\\ y &= -1 \end{align}$$

Double nice! We've figured out the values of $x$ and $y$.

It's really important to check your work . Plug the values of your variables into BOTH equations to make sure they're correct.

Let's try a problem where you make your own system of equations.

John and Joseph are having a competition. John thinks the train is the cheapest way to travel and Joseph thinks airplane is the cheapest way to travel. The total cost of the train and airplane tickets are $\$150$. Joseph paid $\$90$ more than John for his ticket. How much did each person pay?

Write a system of equations using this information. Use the substitution method to solve.

We start to solve this equation by assigning variables to either the train cost or plane cost. Let's say $p$ represents the amount of money a plane ticket costs and $t$ is the amount of money a train ticket costs. Now, we see that the total cost of both tickets was $\$150$. This is modeled by the equation $$p+t=150$$ We know that Joseph's plane ticket is $\$90$ more than the train ticket. We can model that with the equation $$p=t+90$$ Now, we use substitution to solve.

Since $p=t+90,$ we can substitute $t+90$ for $p$ in the first equation, $p+t=150$. We see, $$\begin{align} p+t&=150\\ (t+90)+t&=150\\ 2t+90&=150\\ 2t+90-90&=150-90\\ 2t&=60\\ t&=30 \end{align}$$

Now, we can substitute that $t$ value into any equation and solve for $p$. We see, $$\begin{align} p&=t+90\\ p&=30+90\\ p&=120 \end{align}$$

So, Joseph spent $\$120$ on his plane ticket and John spent $\$30$ on his train ticket. Clearly, the train is cheaper!

Try to check if these solutions are correct by yourself! Remember, substitute both solutions into both equations!

(Video) The Decision - Math Video Challenge

by PBSQ with the MATHCOUNTS Foundation

Let's warm up with the winning video from the 2015 MATHCOUNTS Math Video Challenge!

This video by PBSQ with the MATHCOUNTS Foundation covers the substitution method.

$\large{\text{Summary}}$

Solving systems of equations can be really important, especially when your colleagues have a super weird way of keeping track of basketball statistics. In this video, a college basketball scout needs some help figuring out how many points his prospective players really scored. Eventually, he does it by using the substitution method!

Here is the word problem from the video:

"In a basketball game, Alex and Brock scored $23$ points in all, Alex and Cash scored $21$ points in all, and Brock and Cash scored $20$ points in all. How many points did the three guys score together?"

We can write three equations to model this word problem! If we let $A=\text{Alex's points}, B=\text{Brock's points},$ and $C=\text{Cash's points},$ then:

$$\begin{align} A+B&=23\\ A+C&=21\\ B+C&=20 \end{align}$$

How can we figure out how many points the guys scored all together?

Here are the steps our mathematician used $($the explanation of systems of equations starts at $2:43):$

Step One: Solve for $A$ in two equations.

$$\begin{align} A+B&=23\\ A&=23-B \end{align}$$ $$\begin{align} A+C&=21\\ A&=21-C \end{align}$$

Step Two: Since $23-B$ and $21-C$ are both equal to $A,$ we can set them equal to each other by the transitive property of equality. Then we can solve for $C$ in terms of $B$ (or $B$ in terms of $C,$ if you want).

$$\begin{align} 21-C&=23-B\\ -C&=2-B\\ C&=B-2 \end{align}$$

Step Three: Substitute $C=B-2$ into the equation $B+C=20$ so that we have an equation with just $B's.$ This way, we can solve for $B!$

$$\begin{align} B+C&=20\\ B+(B-2)&=20\\ B+B-2&=20\\ 2B-2&=20\\ 2B&=22\\ B&=\boxed{11} \end{align}$$

Step Four: Now, we can substitute $B=11$ into the equations $A+B=23$ and $B+C=20$ to solve for $A$ and $C!$

$$\begin{align} A+B&= 23\\ A+(11)&=23\\ A&=\boxed{12} \end{align}$$ $$\begin{align} B+C&= 20\\ (11)+C&=20\\ C&=\boxed{9} \end{align}$$

Step Five: Check that your solutions actually work!

$$\begin{align} A+B&=23\\ 12+11&=23\\ 23&=23 \end{align}$$ $$\begin{align} A+C&=21\\ 12+9&=21\\ 21&=21 \end{align}$$ $$\begin{align} B+C&=20\\ 11+9&=20\\ 20&=20 \end{align}$$

Step Six: Our initial word problem asked us to find the total number of points that the guys scored together. To do this, we just have to add together Alex's points, Brock's points, and Cash's points!

$$\begin{align} A+B+C &=12+11+9\\ &=\boxed{32} \end{align}$$

So, in total, the guys scored $32$ points!

Another Example Problem

Now let's try a different word problem.

Solution

Here's the word problem again:

"In a basketball game, Zoey and Yasmine scored $25$ points in all, Yasmine and Xariya scored $31$ points in all, and Xariya and Zoey scored $26$ points in all. How many points did the three gals score together?"

Let's let $Z=\text{Zoey's points}, Y=\text{Yasmine's points},$ and $X=\text{Xariya's points}.$ Now we have the following system of equations:

$$\begin{align} Z+Y&=25\\ Y+X&=31\\ X+Z&=26 \end{align}$$

Step One: Solve for a variable in two equations. It can be any of the variables, so let's just pick $Z.$

$$\begin{align} Z+Y&=25\\ Z&=25-Y \end{align}$$ $$\begin{align} X+Z&=26\\ Z&=26-X \end{align}$$

Step Two: Since $25-Y$ and $26-X$ are both equal to $Z,$ we can set them equal to each other. Then we can solve for $X$ in terms of $Y$ (or $Y$ in terms of $X,$ if you want).

$$\begin{align} 26-X&=25-Y\\ -X&=-1-Y\\ X&=Y+1 \end{align}$$

Step Three: Substitute $X=Y+1$ into the equation $Y+X=31$ so that we have an equation with just $Y's.$ This way, we can solve for $Y!$

$$\begin{align} Y+X&=31\\ Y+(Y+1)&=31\\ Y+Y+1&=31\\ 2Y+1&=31\\ 2Y&=30\\ Y&=\boxed{15} \end{align}$$

Step Four: Now, we can substitute $Y=15$ into the equations $Z+Y=25$ and $Y+X=31$ to solve for $Z$ and $X!$

$$\begin{align} Z+Y&=25\\ Z+(15)&=25\\ Z&=\boxed{10} \end{align}$$ $$\begin{align} Y+X&=31\\ (15)+X&=31\\ X&=\boxed{16} \end{align}$$

Step Five: Check that your solutions actually work!

$$\begin{align} Z+Y&=25\\ 10+15&=25\\ 25&=25 \end{align}$$ $$\begin{align} Y+X&=31\\ 15+6&=31\\ 31&=31 \end{align}$$ $$\begin{align} X+Z&=26\\ 16+10&=26\\ 26&=26 \end{align}$$

Step Six: Our initial word problem asked us to find the total number of points that the gals scored together. To do this, we just have to add together Zoey's points, Yasmine's points, and Xariya's points!

$$\begin{align} Z+Y+X &=10+15+16\\ &=\boxed{41} \end{align}$$

So, in total, the gals scored $41$ points!

Earlier, we saw how to solve a system of linear equations by graphing. Later, you will look at solving systems via elimination. After you master graphing, substitution, and elimination, we'll tackle word problems with systems of equations!

Here, we will use the substitution method to solve a system of linear equations. It is used to eliminate one variable by substituting the other in a system of equations.

Let's take a look at an example: $$\begin{align} x-3y&=-10\\ 3x+2y&=14 \end{align}$$

Step One

Single out one of the variables by solving one of the equations for $x$ or $y.$ Here, let's solve for $x$ in the first equation. $$\begin{align} x-3y&=-10\\ x &= -10 + 3y \end{align}$$

Step Two

Substitute the found variable or expression into the second equation, solving the resulting equation. $$\begin{align} 3x+2y&=14\\ 3(-10+3y)+2y&=14, \text{ since we know } x=-10+3y \text{ from Step One} \end{align}$$

Step Three

Solve this new linear equation. In this example, we solve for $y$. $$\begin{align} 3(-10+3y)+2y&=14 \\ -30+9y+2y &= 14 \\ 9y+2y &= 44 \\ 11y& = 44 \\\ y& = 4 \end{align}$$

Step Four

Plug the value of the variable you just solved for into either of the original equations to solve for the first variable. Here, let's substitute $y=4$ back into the first equation. $$\begin{align} x-3y&=-10\\ x -3(4) &=-10\\ x-12 &=-10\\ x&=2 \end{align}$$

Step Five

Double check! Substitute the values of $x$ and $y$ (the ordered pair you found) into the original equations. If both sides of both equations are equal, then well done! $$\begin{align} 3x+2y&=14 \\ 3(2)+2(4)&=14\\ 6+8&=14\\ 14&=14 \end{align}$$

$$\begin{align} x-3y&=-10 \\ (2)-3(4)&=-10\\ 2-12&=-10\\ -10&=-10 \end{align}$$

Here's a graphic with this example!

A system of linear equations can be written and solved using matrix–vector multiplication. You'll learn that soon!