Course: 3.9 双方有变量的等同

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两边都有变量的公式
Equations with Variables on Both Sides
::两边都有变量的公式

When a variable appears on both sides of the equation , we need to manipulate the equation so that all variable terms appear on one side, and only constants are left on the other.
::当一个变量出现在方程的两侧时,我们需要操纵方程,以便所有变量条件都出现在一边,而只有常数留在另一侧。

Real-World Application: Chemistry
::现实世界应用:化学

Dwayne was told by his chemistry teacher to measure the weight of an empty beaker using a balance. Dwayne found only one lb weights, and so devised the following way of balancing the scales.
::Dwayne的化学老师告诉Dwayne用平衡测量空瓶子的重量。 Dwayne只发现了一磅重量,因此设计了平衡比例的以下方法。

Knowing that each weight is one lb, calculate the weight of one beaker.
::知道每个重量是一磅, 计算一个烧杯的重量。

We know that the system balances, so the weights on each side must be equal. We can write an algebraic expression based on this fact. The unknown quantity, the weight of the beaker, will be our $\begin{aligned} x \end{aligned}$ . We can see that on the left hand scale we have one beaker and four weights. On the right scale, we have four beakers and three weights. The balancing of the scales is analogous to the balancing of the following equation:
::我们知道系统平衡, 所以每边的重量必须相等。我们可以根据这个事实写一个代数表达式。未知的数量, 烧杯的重量, 将是我们的x。我们可以看到, 在左手秤上, 我们有一个烧杯和四个重量。在右手秤上, 我们有四个烧杯和三个重量。比例的平衡与以下等式的平衡相似:

$\begin{aligned} x + 4 = 4 x + 3 \end{aligned}$

::x+4=4x+3 x+4=4x+3

“One beaker plus 4 lbs equals 4 beakers plus 3 lbs”
::“一杯加4磅等于4杯加3磅”

To solve for the weight of the beaker, we want all the constants (numbers) on one side and all the variables (terms with $\begin{aligned} x \end{aligned}$ in them) on the other side. Since there are more beakers on the right and more weights on the left, we’ll try to move all the $\begin{aligned} x \end{aligned}$ terms (beakers) to the right, and the constants (weights) to the left.
::为了解决烧杯的重量问题,我们需要一面的所有常数(数字)和另一侧的所有变量(用x来配制的值)。由于右边有更多的烧杯,左侧的重则更多,我们将尝试将所有x条件(水杯)移到右边,将常数(重量)移到左边。

First we subtract 3 from both sides to get $\begin{aligned} x + 1 = 4 x \end{aligned}$ .
::首先我们从两边减去3 来获得 x+1=4x 。

Then we subtract $\begin{aligned} x \end{aligned}$ from both sides to get $\begin{aligned} 1 = 3 x \end{aligned}$ .
::然后我们从两边减去 x 以获得 1=3x 。

Finally we divide by 3 to get $\begin{aligned} \frac{1}{3} = x \end{aligned}$ .
::最后我们除以3 13=x。

The weight of the beaker is one-third of a pound.
::烧杯的重量是一磅的三分之一

We can do the same with the real objects as we did with the equation. Just as we subtracted amounts from each side of the equation, we could remove a certain number of weights or beakers from each scale. Because we remove the same number of objects from each side, we know the scales will still balance.
::我们可以对真实物体和方程一样做同样的事情。正如我们从方程的每侧减去数量一样,我们可以从每个尺度中去除一定数量的重量或烧杯。由于我们从每一边删除了同样数量的物体,我们知道天平仍然会保持平衡。

First, we could remove three weights from each scale. This would leave one beaker and one weight on the left and four beakers on the right (in other words $\begin{aligned} x + 1 = 4 x \end{aligned}$ ):
::首先,我们可以从每个尺度中去除三个重量。这样,左边将留下一个烧杯,左边将留下一个重量,右边将留下四个烧杯(换句话说,X+1=4x):

Then we could remove one beaker from each scale, leaving only one weight on the left and three beakers on the right, to get $\begin{aligned} 1 = 3 x \end{aligned}$ :
::然后,我们可以从每个尺度中取出一个杯子, 左边只留一个重量,右边留三个杯子, 以得到1=3x:

Looking at the balance, it is clear that the weight of one beaker is one-third of a pound.
::审视平衡情况,显然,1个烧杯的重量是1磅的三分之一。

Solve an Equation with Grouping Symbols
::用群組符号解析等式

As you’ve seen, we can solve equations with variables on both sides even when some of the variables are in " data-term="Parentheses" role="term" tabindex="0"> parentheses ; we just have to get rid of the parentheses, and then we can start combining like terms . We use the same technique when dealing with fractions: first we multiply to get rid of the fractions, and then we can shuffle the terms around by adding and subtracting.
::正如你们所看到的,即使有些变量在括号中,我们也能用两边的变量解决方程式问题;我们只需要删除括号,然后我们就可以像用词一样合并。我们在处理分数时使用同样的技术:首先,我们乘法来消除分数,然后,我们可以通过增减来冲洗两边的术语。

Solving for Unknown Values
::解决未知值

1. Solve $\begin{aligned} 3 x + 2 = \frac{5 x}{3} \end{aligned}$ .
::1. 解决 3x+2=5x3。

The first thing we’ll do is get rid of the fraction . We can do this by multiplying both sides by 3, leaving $\begin{aligned} 3 (3 x + 2) = 5 x \end{aligned}$ .
::我们要做的第一件事就是除掉这个分数。我们可以通过将两边乘以3来做到这一点,留下3(3)(3x+2)=5x。

Then we distribute to get rid of the parentheses, leaving $\begin{aligned} 9 x + 6 = 5 x \end{aligned}$ .
::然后我们分发去掉括号留下9x6+6=5x

We’ve already got all the constants on the left side, so we’ll move the variables to the right side by subtracting $\begin{aligned} 9 x \end{aligned}$ from both sides. That leaves us with $\begin{aligned} 6 = - 4 x \end{aligned}$ .
::我们已经在左侧得到了所有常数,所以我们将将变量移到右侧,从两侧减去9x。这就只剩下6+4x了。

And finally, we divide by -4 to get $\begin{aligned} - \frac{3}{2} = x \end{aligned}$ , or $\begin{aligned} x = - 1.5 \end{aligned}$ .
::最后,我们除以 -4 以获得 - 32=x, 或 x1.5。

2. Solve the following equation for $\begin{aligned} x \end{aligned}$ : $\begin{aligned} \frac{14 x}{(x + 3)} = 7 \end{aligned}$
::2. 解决xx的下列方程: 14x(x+3)=7

The form of the left hand side of this equation is known as a rational function because it is the ratio of two other functions: $\begin{aligned} 14 x \end{aligned}$ and $\begin{aligned} (x + 3) \end{aligned}$ . But we can solve it just like any other equation involving fractions.
::此方程式的左侧形式被称为理性函数, 因为它是另外两个函数( 14x 和 (x+3)) 的比, 但我们可以像任何其他包含分数的方程式一样解决它。

First we multiply both sides by $\begin{aligned} (x + 3) \end{aligned}$ to get rid of the fraction. Now our equation is $\begin{aligned} 14 x = 7 (x + 3) \end{aligned}$ .
::首先我们要将两边乘以 (x+3) 来清除分数。现在我们的方程式是 14x= 7(x+3) 。

Then we distribute: $\begin{aligned} 14 x = 7 x + 21 \end{aligned}$ .
::然后我们分发: 14x=7x+21。

Then subtract $\begin{aligned} 7 x \end{aligned}$ from both sides: $\begin{aligned} 7 x = 21 \end{aligned}$ .
::然后从两边减去7x7:7x=21。

And divide by 7: $\begin{aligned} x = 3 \end{aligned}$ .
::除以7: x=3,除以7: x=3。

Solve Real-World Problems Using Equations with Variables on Both Sides
::使用两边变量的等式解决现实世界问题

Here’s another chance to practice translating problems from words to equations. What is the equation asking? What is the unknown variable? What quantity will we use for our variable?
::这是将问题从文字转换为方程式的另一个机会。方程式要问什么? 未知变量是什么? 我们的变量要用多少量?

The text explains what’s happening. Break it down into small, manageable chunks, and follow what’s going on with our variable all the way through the problem.
::文本解释了正在发生的事情。将它分为小块、可控的块块,并跟踪我们变数的动态,一直到问题解决为止。

More on Ohm’s Law
::更多关于Ohm法

Recall that the electrical current, $\begin{aligned} I \end{aligned}$ (amps), passing through an electronic component varies directly with the applied voltage, $\begin{aligned} V \end{aligned}$ (volts), according to the relationship $\begin{aligned} V = I \cdot R \end{aligned}$ where $\begin{aligned} R \end{aligned}$ is the resistance measured in Ohms $\begin{aligned} (Ω) \end{aligned}$ .
::回顾电流I(装置)通过电子部件,与应用电压V(电压)直接不同,根据V=IR的关系,R是用Ohms(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

The resistance $\begin{aligned} R \end{aligned}$ of a number of components wired in a series (one after the other) is simply the sum of all the resistances of the individual components.
::一系列(一个接一个接一个接一个的)电线的若干部件的抗力R只是每个部件所有阻力的总和。

In an attempt to find the resistance of a new component, a scientist tests it in series with standard resistors. A fixed voltage causes a 4.8 amp current in a circuit made up from the new component plus a $\begin{aligned} 15 Ω \end{aligned}$ resistor in series. When the component is placed in a series circuit with a $\begin{aligned} 50 Ω \end{aligned}$ resistor, the same voltage causes a 2.0 amp current to flow. Calculate the resistance of the new component.
::为了寻找新元件的抗力,科学家用标准阻力器进行系列试验。固定电压在由新元件组成的电路中产生4.8个脉冲电流,加上1 15-38个阻力器序列。当该元件被放置在一个配有50-38抗力器的系列电路中时,同样的电压导致2.0个脉冲流。计算新元件的阻力。

This is a complex problem to translate, but once we convert the information into equations it’s relatively straightforward to solve. First, we are trying to find the resistance of the new component (in Ohms, $\begin{aligned} Ω \end{aligned}$ ). This is our $\begin{aligned} x \end{aligned}$ . We don’t know the voltage that is being used, but we can leave that as a variable, $\begin{aligned} V \end{aligned}$ . Our first situation has a total resistance that equals the unknown resistance plus $\begin{aligned} 15 Ω \end{aligned}$ . The current is 4.8 amps. Substituting into the formula $\begin{aligned} V = I \cdot R \end{aligned}$ , we get $\begin{aligned} V = 4.8 (x + 15) \end{aligned}$ .
::这是一个复杂的翻译问题,但一旦我们把信息转换成方程式,它就比较容易解决。首先,我们试图找到新组件的阻力(在 Ohms, _)。这是我们的x。我们不知道正在使用的电压,但是我们可以将它作为一个变量,V。我们的第一个情况是完全阻力,等于未知阻力加上15/38。当前是4.8安培。我们用V=IQR,我们得到V=4.8(x+15)。

Our second situation has a total resistance that equals the unknown resistance plus $\begin{aligned} 50 Ω \end{aligned}$ . The current is 2.0 amps. Substituting into the same equation, this time we get $\begin{aligned} V = 2 (x + 50) \end{aligned}$ .
::我们的第二个情况是完全的抵抗力等于未知的抵抗力加上50/ 28。目前的是2.0 AMPs。以相同的方程替代, 这次我们得到 V=2( x+50) 。

We know the voltage is fixed, so the $\begin{aligned} V \end{aligned}$ in the first equation must equal the $\begin{aligned} V \end{aligned}$ in the second. That means we can set the right-hand sides of the two equations equal to each other: $\begin{aligned} 4.8 (x + 15) = 2 (x + 50) \end{aligned}$ . Then we can solve for $\begin{aligned} x \end{aligned}$ .
::我们知道电压是固定的, 所以第一个方程式中的V必须等于第二个方程式中的V。这意味着我们可以设置两个方程式的右侧, 即4. 8( x+15)=2( x+50)。然后我们就可以解决 x 。

Distribute the constants first: $\begin{aligned} 4.8 x + 72 = 2 x + 100 \end{aligned}$ .
::先分配常数: 4. 8x+72=2x+100。

Subtract $\begin{aligned} 2 x \end{aligned}$ from both sides: $\begin{aligned} 2.8 x + 72 = 100 \end{aligned}$ .
::双方减号2x:2.8x+72=100。

Subtract 72 from both sides: $\begin{aligned} 2.8 x = 28 \end{aligned}$ .
::双方减72:2.8x=28。

Divide by 2.8: $\begin{aligned} x = 10 \end{aligned}$ .
::除以2.8:x=10。

The resistance of the component is $\begin{aligned} 10 Ω \end{aligned}$ .
::部件的阻力是10进30分

Examples
::实例

Example 1
::例1

Sven was told to find the weight of an empty box with a balance. Sven found some one lb weights and five lb weights. He placed two one lb weights in three of the boxes and with a fourth empty box found the following way of balancing the scales:
::Sven被告知找到一个有平衡的空箱的重量。Sven发现了一些一磅重量和五磅重量。他把两个一磅重量放在三个箱子中,而第四个空箱发现平衡比例的下列方法:

Knowing that small weights are one lb and big weights are five lbs, calculate the weight of one box.
::知道小体重是一磅,大体重是五磅, 计算一个盒子的重量。

We know that the system balances, so the weights on each side must be equal. We can write an algebraic expression based on this equality. The unknown quantity—the weight of each empty box, in pounds—will be our $\begin{aligned} x \end{aligned}$ . A box with two 1 lb weights in it weighs $\begin{aligned} (x + 2) \end{aligned}$ pounds. Our equation, based on the picture, is $\begin{aligned} 3 (x + 2) = x + 3 (5) \end{aligned}$ .
::我们知道系统平衡, 所以每边的重量必须是相等的。我们可以根据这种平等写一个代数表达式。未知的数量—— 每个空箱的重量, 磅—— 将是我们的。一个重量为两磅的盒子。我们基于图片的方程式是 3 (x+2)=x+3(5)。

Distributing the 3 and simplifying, we get $\begin{aligned} 3 x + 6 = x + 15 \end{aligned}$ .
::分发3和简化, 我们得到3x+6=x+15。

Subtracting $\begin{aligned} x \end{aligned}$ from both sides, we get $\begin{aligned} 2 x + 6 = 15 \end{aligned}$ .
::从两边减去 x, 我们得到 2x+6=15。

Subtracting 6 from both sides, we get $\begin{aligned} 2 x = 9 \end{aligned}$ .
::从两边减去6, 我们得到2x=9。

And finally we can divide by 2 to get $\begin{aligned} x = \frac{9}{2} \end{aligned}$ , or $\begin{aligned} x = 4.5 \end{aligned}$ .
::最终我们可以除以 2 以获得 x=92, 或 x=4.5。

Each box weighs 4.5 lbs.

::每个盒子重4.5磅。

Example 2
::例2

Solve $\begin{aligned} 7 x + 2 = \frac{5 x - 3}{6} \end{aligned}$ .
::解决 7x+2=5x-36。

Again we start by eliminating the fraction. Multiplying both sides by 6 gives us $\begin{aligned} 6 (7 x + 2) = 5 x - 3 \end{aligned}$ , and distributing gives us $\begin{aligned} 42 x + 12 = 5 x - 3 \end{aligned}$ .
::我们再次从消除分数开始。将两边乘以 6 给我们 6( 7x+2) = 5x- 3, 分配给我们 42x+12= 5x- 3 。

Subtracting $\begin{aligned} 5 x \end{aligned}$ from both sides gives us $\begin{aligned} 37 x + 12 = - 3 \end{aligned}$ .
::从两边减5x 共37x+12+3

Subtracting 12 from both sides gives us $\begin{aligned} 37 x = - 15 \end{aligned}$ .
::从两边减12,我们37x15。

Finally, dividing by 37 gives us $\begin{aligned} x = - \frac{15}{37} \end{aligned}$ .
::最后,以37除以37 给了我们1x1537

Review
::回顾

For 1-11, solve the following equations for the unknown variable.
::1- 11, 解析未知变量的下列方程式。

$\begin{aligned} 3 (x - 1) = 2 (x + 3) \end{aligned}$
::3(x-1)=2(x+3)

$\begin{aligned} 7 (x + 20) = x + 5 \end{aligned}$
::7(x+20)=x+5

$\begin{aligned} 9 (x - 2) = 3 x + 3 \end{aligned}$
::9(x-2)=3x+3

$\begin{aligned} 2 (a - \frac{1}{3}) = \frac{2}{5} (a + \frac{2}{3}) \end{aligned}$
::2(a)-13)=25(a+23)

$\begin{aligned} \frac{2}{7} (t + \frac{2}{3}) = \frac{1}{5} (t - \frac{2}{3}) \end{aligned}$
::27(t+23)=15(t-23)

$\begin{aligned} \frac{1}{7} (v + \frac{1}{4}) = 2 (\frac{3 v}{2} - \frac{5}{2}) \end{aligned}$
::17(v+14)=2(3v2-2-52)

$\begin{aligned} \frac{y - 4}{11} = \frac{2}{5} \cdot \frac{2 y + 1}{3} \end{aligned}$
::y-411=252y+13

$\begin{aligned} \frac{z}{16} = \frac{2 (3 z + 1)}{9} \end{aligned}$
::z16=2(3z+1)9

$\begin{aligned} \frac{q}{16} + \frac{q}{6} = \frac{(3 q + 1)}{9} + \frac{3}{2} \end{aligned}$
::q16+q6=(3q+1)9+32

$\begin{aligned} \frac{3}{x} = \frac{2}{x + 1} \end{aligned}$
::3x=2x+1

$\begin{aligned} \frac{5}{2 + p} = \frac{3}{p - 8} \end{aligned}$
::52+p=3p-8

Manoj and Tamar are arguing about a number trick they heard. Tamar tells Andrew to think of a number, multiply it by five and subtract three from the result. Then Manoj tells Andrew to think of a number, add five and multiply the result by three. Andrew says that whichever way he does the trick he gets the same answer.

What was the number Andrew started with?
::安德鲁的号码是多少?

What was the result Andrew got both times?
::安德鲁两次得到的结果是什么?

Name another set of steps that would have resulted in the same answer if Andrew started with the same number.
::列出另一套步骤,如果安德鲁以同样数字开始,将得出同样的答案。

::Manoj 和 Tamar 正在争论他们听到的数个骗局。 Tamar 告诉 Andrew 想想一个数字, 乘以5, 从结果中减去3。然后Manoj 告诉 Andrew 想想一个数字, 加上5, 乘以3。 Andrew 说无论他用什么方式玩这个把戏, 他得到的答案是一样的。 Andrew 一开始用什么数字? Andrew 两次都得到了什么结果? 说出另一套步骤,如果Andrew 以同一个数字开始, 就会得出同样的答案。

Manoj and Tamar try to come up with a harder trick. Manoj tells Andrew to think of a number, double it, add six, and then divide the result by two. Tamar tells Andrew to think of a number, add five, triple the result, subtract six, and then divide the result by three.

Andrew tries the trick both ways and gets an answer of 10 each time. What number did he start out with?
::安德鲁尝试两招每次都得到10个答案

He tries again and gets 2 both times. What number did he start out with?
::他又试了一次,两次得到两次他从哪个数字开始?

Is there a number Andrew can start with that will not give him the same answer both ways?
::安德鲁能不能从这个开始给他两个不同的答案?

Bonus: Name another set of steps that would give Andrew the same answer every time as he would get from Manoj’s and Tamar’s steps.
::福利: 说出另一套步骤, 每次安德鲁从Manoj和Tamar的脚步中得到同样的答案,

::Manoj 和 Tamar 尝试用一个更难的把戏。 Manoj 告诉 Andrew 想到一个数字, 加倍它, 增加6, 然后将结果除以2。 Tamar 告诉 Andrew 想到一个数字, 增加5, 三倍结果, 减去6, 然后将结果除以3。 Andrew 尝试了两种把戏, 每次得到10个答案。他从哪个数字开始? 他又尝试了两个数字。他从哪个数字开始? Andrew 能从哪个数字开始, 给他两个答案都不相同吗? Bonus : 列出另一套步骤, 每从Manoj 和 Tamar 的阶梯上给Andrew同样的答案。

I have enough money to buy five regular priced CDs and have $6 left over. However, all CDs are on sale today, for $4 less than usual. If I borrow $2, I can afford nine of them.

How much are CDs on sale for today?
::今天的CD卖多少钱?

How much would I have to borrow to afford nine of them if they weren’t on sale?
::买得起九张票要借多少钱?

::我有足够的钱购买五张固定价格的CD,还有6美元剩余。然而,今天所有CD都在出售,比平常少4美元。如果我借2美元,我可以买9张。今天的CD要卖多少钱? 如果不卖,我还要借多少才能买9张?

Five identical electronics components were connected in series. A fixed but unknown voltage placed across them caused a 2.3 amp current to flow. When two of the components were replaced with standard $\begin{aligned} 10 Ω \end{aligned}$ resistors, the current dropped to 1.9 amps. What is the resistance of each component?
::五个相同的电子部件连成系列连接。一个固定但未知的电压在它们之间造成2.3个脉冲流。当其中两个部件被标准10(8)阻力器取代时,电流下降为1.9安普。每个部件的阻力是什么?

Solve the following resistance problems. Assume the same voltage is applied to all circuits.

Three unknown resistors plus $\begin{aligned} 20 Ω \end{aligned}$ give the same current as one unknown resistor plus $\begin{aligned} 70 Ω \end{aligned}$ .
::3个未知的抵抗者加上20年与1个未知的抵抗者加上70年

One unknown resistor gives a current of 1.5 amps and a $\begin{aligned} 15 Ω \end{aligned}$ resistor gives a current of 3.0 amps.
::一个未知的阻力器给出了1.5安普斯的电流,一个15-38的阻力器给出了3.0安普斯的电流。

Seven unknown resistors plus $\begin{aligned} 18 Ω \end{aligned}$ gives twice the current of two unknown resistors plus $\begin{aligned} 150 Ω \end{aligned}$ .
::7个未知的抵抗者加上1828年给了2个未知的抵抗者两倍的电流加上150年的电流

Three unknown resistors plus $\begin{aligned} 1.5 Ω \end{aligned}$ gives a current of 3.6 amps and seven unknown resistors plus seven $\begin{aligned} 12 Ω \end{aligned}$ resistors gives a current of 0.2 amps.
::3个未知阻力器加上1.5和1.5的电流为3.6个安培器,7个未知阻力器加上7个12和12的阻力器的电流为0.2个安培器。

::解决以下抗力问题。对所有电路都使用相同的电压。 3个未知抗体加上20/ 3个未知阻力器加上20/ 3个未知阻力器加上20/ 3个未知阻力器加70/ 70/ 3个未知阻力器加上1个未知阻力器加70/ 3个未知阻力器,1个1个未知阻力器提供1.5个安培器,1个15/ 3个阻力器提供3. 0个安培器。 7个未知阻力器加上18/ 3个未知的阻力加上2个未知阻力加上150/ 3个未知的阻力加上1.5/ 3个未知阻力器加7个未知阻力器加上7个12/ 3个未知阻力器给予0. 2个安培力器的电流。

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3.9 双方有变量的等同

Section outline

General

两边都有变量的公式

Equations with Variables on Both Sides
::两边都有变量的公式

Real-World Application: Chemistry
::现实世界应用:化学

Solving for Unknown Values
::解决未知值

Solve Real-World Problems Using Equations with Variables on Both Sides
::使用两边变量的等式解决现实世界问题

More on Ohm’s Law
::更多关于Ohm法

Examples
::实例

Example 1
::例1

Example 2
::例2

Review
::回顾

Review (Answers)
::回顾(答复)

Section outline

General

两边都有变量的公式

Equations with Variables on Both Sides ::两边都有变量的公式

Real-World Application: Chemistry ::现实世界应用:化学

Solving for Unknown Values ::解决未知值

Solve Real-World Problems Using Equations with Variables on Both Sides ::使用两边变量的等式解决现实世界问题

More on Ohm’s Law ::更多关于Ohm法

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Review ::回顾

Review (Answers) ::回顾(答复)

Equations with Variables on Both Sides
::两边都有变量的公式

Real-World Application: Chemistry
::现实世界应用:化学

Solving for Unknown Values
::解决未知值

Solve Real-World Problems Using Equations with Variables on Both Sides
::使用两边变量的等式解决现实世界问题

More on Ohm’s Law
::更多关于Ohm法

Examples
::实例

Example 1
::例1

Example 2
::例2

Review
::回顾

Review (Answers)
::回顾(答复)