Expressions and Equations

Remember, what we’re looking for are “numbers and variables joined together with math”, and we want to make sure we don’t have any equal signs. That makes options a and d examples of algebraic expressions. Examples b and c aren’t algebraic expressions, because they each include an equal sign!

We’ll take this an expression at a time.

1. When we look at $6 + x \cdot \frac{1}{2} - 3y$, it’s almost a rewrite of division, but it’s missing a key feature: parentheses. We want to make sure we’re multiplying the entire numerator by the reciprocal of the entire denominator. When we write $6 + x \cdot \frac{1}{2}$, only the $x$ is being multiplied by the $\frac{1}{2}$, not the entire $6 + x$. To fix this, then, our expression should be $(6 + x) \cdot \frac{1}{2} - 3y$. Now it’s equivalent to the original.

2. Next we consider $\frac{x + 6}{2} - 3y$. The only thing that’s happened here is that the addition in the numerator has been rearranged, a perfectly valid move. Therefore, this expression is equivalent to the original.

3. In $3y - \frac{x + 6}{2}$, we swapped around the subtraction. The problem is, subtraction isn’t commutative, so this is not equivalent to the original expression.

4. Now this is more like it. If we first change $\frac{6 + x}{2} - 3y$ into $\frac{6 + x}{2} + ( - 3y)$, then we have some addition that we can rearrange. So, we rewrite as $- 3y + \frac{6 + x}{2}$ for an equivalent expression.

5. Check out the extra 4’s in $\frac{4(6 + x)}{4 \cdot 2} - 3y$. This comes from multiplying that first fraction by $\frac{4}{4}$, a fancy form of 1, so this expression is still equivalent to the original.

6. In $\frac{6 + x}{2} + 5 - 3y$, we simply added a 5 to the expression. That definitely changes the value of the expression, so it is not equivalent to the original.

7. This last expression can be created if we add $5 - 5$ to the original expression. Is that allowed? Sure, because $5 - 5$ is a fancy form of 0. So we have
   $\frac{6 + x}{2} - 3y + 5 - 5$, which then allows us to (carefully) rearrange our addition to get $\frac{6 + x}{2} + 5 - 3y - 5.$

1. Check out the operations in $5(3x)$. That’s 5 times 3 times $x$; there’s no addition or subtraction in sight, so distribution doesn’t apply here. To simplify, we can think of this instead as $5 \cdot 3 \cdot x$, so $15x$.

2. Now we have 5 times $3 + x$, so we have the required multiplication by an addition expression. So, we distribute the 5:
   \[{5 \cdot 3 + 5 \cdot x }{= 15 + 5x}\]

3. Distribution still applies when we have a number (5) times a subtraction expression $(3 - x)$, so to simplify we can distribute the 5:
   \[{5 \cdot 3 - 5 \cdot x }{= 15 - 5x}\]

4. Now we have an addition expression, $3 + x$, raised to a power of 5. That’s not multiplication, so distribution doesn’t apply here. It would be incorrect to write $3^{5} + x^{5}$. In fact, if we wanted to simplify, we’d need to think about $(3 + x)$ multiplied by itself 5 times, like $(3 + x)(3 + x)(3 + x)(3 + x)(3 + x)$.

5. We in fact do have multiplication meeting with addition here, so we can distribute…carefully. We can think of this as $x + 5$ times an addition statement, $x + 3$. So we’ll distribute the $x + 5$ itself:
   \[(x + 5) \cdot x + (x + 5) \cdot 3 \]We can rearrange the multiplication a little, then distribute again:
   \[{x(x + 5) + 3(x + 5) }{= x \cdot x + x \cdot 5 + 3 \cdot x + 3 \cdot 5}\]

\[= x^{2} + 5x + 3x + 15 \]This expression can be simplified even further, but first we must discuss some more simplification methods!

Only item b is not a term. The 3 and $x$ are joined with addition, not multiplication. In fact, the 3 and $x$ are each an individual term!

The like terms are $3x^{2}$ and $5x^{2}$, because they both have the same variable ($x$) raised to the same power $(2)$. In the $3x^{2}$ term, 3 is the coefficient. Similarly in the $5x^{2}$ term, the 5 is the coefficient. Notice that $3x$ is not a like term in this group, because though it has an $x,\ $that $x$ is raised to the (invisible) first power.

The $3x^{2}y\ $and $7x^{2}y$ are also like terms, because they both have the same combination of variables ($x$ and $y$) each raised to the same power $(2\ $and 1, respectively). Notice that $2xy^{2}$ is not a like term in this group, because its $x$ is raised to the 1^st power and its $y$ is raised to the 2^nd power; it has the same variables, but the wrong powers on those variables.

Let’s start by identifying our like terms. We have $3xy^{2}$ and $2xy^{2}$ for one group, and $7x^{2}y$ and $- x^{2}y$ for our other group. If we change that subtraction to adding the opposite, we can rearrange all the addition to put the like terms next to each other.

\[3xy^{2} + 7x^{2}y + 2xy^{2} + ( - x^{2}y)\]

\[= 3xy^{2} + 2xy^{2} + 7x^{2}y + ( - x^{2}y)\]

Now for the combining part. We’re basically counting up how many of each like term we have. So there are 3 $xy^{2}$’s plus another 2 of the $xy^{2}$’s , making a total of $5xy^{2}$. Similarly, we have 7 of the $x^{2}y$’s, plus a negative 1 $x^{2}y$, so there are $6\ $total $x^{2}y's$. This means, we have

\[5xy^{2} + 6x^{2}y\]

for our final simplified answer.

Now we have some distribution and eventually some like terms to combine. To ensure that the negatives don’t mess us up, though, let’s change that $- 3$ into adding the opposite:

\[5(x - 6) + ( - 3)(x + 4)\]

Now we distribute, rearrange, and combine like terms.

\[5x - 5 \cdot 6 + \ - 3x + \ - 3 \cdot 4\]

\[= 5x + - 30 + \ - 3x + \ - 12\]

\[= 5x + \ - 3x + \ - 30 + \ - 12\]

\[= 2x - 42\]

For each of these, our primary job is to attend to the order of operations.

1. \[\begin{array}{ll} 3 \cdot 4 - (5 - 3)^{2} + 2 & \\ = 3 \cdot 4 - \left( 2 \right)^{2} + 2 & \text{parentheses} \\ = 3 \cdot 4 - 4 + 2 & \text{exponents} \\ = 12 - 4 + 2 & \text{multiplication and division} \\ = 8 + 2 & \text{addition and subtraction} \\ = 10 & \text{addition and subtraction} \end{array}\]

Notice that once this expression simplified down to all addition and subtraction, we didn’t necessarily do addition first, then subtractions. The “AS” in our “PEMDAS” mnemonic tells us to do addition and subtraction from left to right.

2. When we have a big fraction like this, we want to think of the entire numerator and the entire denominator as each being in their own set of parentheses (even if the expression isn’t explicitly written like that).
   \[\begin{array}{ll} \frac{(4 + 2)}{(4 - 1)} & \\ = \frac{6}{3} & \text{parentheses} \\ = 2 & \text{division} \end{array}\]

3. Our friends, fractions! We’ll review some fraction basics while using the order of operations.

First, recall that to add fractions, we need to make sure they have a common denominator. How can we make a common denominator? Multiply our fractions by a fancy form of 1 (one of the things we’re always allowed to do to an expression!).

\[\begin{array}{ll} \frac{1}{2}\left( \frac{3}{4} + \frac{1}{2} \right) & \\ = \frac{1}{2}\left( \frac{3}{4} + \frac{1}{2} \cdot \frac{2}{2} \right) & \text{fancy form of 1} \\ = \frac{1}{2}\left( \frac{3}{4} + \frac{2}{4} \right) & \text{multiplication} \\ = \frac{1}{2}\left( \frac{5}{4} \right) & \text{addition} \\ = \frac{5}{8} & \text{multiplication} \end{array}\]

We want to replace the $x$ and $y$ in our expression with the indicated values. As we do so, it’s a good idea to put our values into parentheses, so we can keep proper track of the operations we’re doing.

1. $x^{2} - 3xy$ where $x = 2, y = 1$

We’ll replace each $x$ with $(2)$ and each $y$ with $(1)$.

\[\begin{array}{ll} x^2 - 3xy & \\ = \left( 2 \right)^{2} - 3\left( 2 \right)\left( 1 \right) & \text{substitute} \\ = 4 - 3(2)(1) & \text{exponents} \\ = 4 - 6(1) & \text{multiplication} \\ = 4 - 6 & \text{multiplication} \\ = - 2 & \text{subtraction} \end{array}\]

2. $x^{2} - 3xy$ where $x = - 2$, $y = 3$

This time, we replace each $x$ with $( - 2)$ and each $y$ with $(3)$.

\[\begin{array}{ll} x^2 - 3xy & \\ = \left( - 2 \right)^2 - 3\left( - 2 \right)\left( 3 \right) & \text{substitute} \\ = 4 - 3( - 2)(3) & \text{exponents}^* \\ = 4+\left( - 3 \right)( - 2)(3) & \text{change subtraction}^\text{**} \\ = 4 + 6(3) & \text{multiplication} \\ = 4 + 18 & \text{multiplication} \\ = 22 & \text{addition} \end{array}\]

*Notice that $( - 2)^{2}\ $is $- 2 \cdot - 2$, hence positive 4.

**Changing the subtraction into adding the opposite helps us to keep proper track of the signs as we work on the multiplying steps.

4. $x^{2} - 3xy$ where $x = z$, $y = z + 2$

Now we have an extra variable floating around, but our process is still the same! We’ll replace each $x$ with $z$ and replace each $y$ with $(z + 2)$.

\[\begin{array}{ll} x^2 - 3xy & \\ = z^2 - 3z\left( z + 2 \right) & \text{substitute} \\ = z^2+\left( - 3z \right)(z + 2) & \text{change subtraction}^* \\ = z^2 + \ - 3z \cdot z + \ - 3z \cdot 2 & \text{parentheses (distribute)}^** \\ = z^2 + \ - 3z \cdot z + \ - 3 \cdot 2 \cdot z & \text{reorder multiplication} \\ = z^2 + \ - 3z^2 + \ - 6z & \text{multiply} \\ = \ - 2z^2 - 6z & \text{combine like terms} \end{array}\]

*Again, we change the subtraction to help keep track of the signs.

**Here, the $z + 2$ in the parentheses can’t be added more, because $z$ and $2$ aren’t like terms. So, instead, we use the distributive property to evaluate the parentheses.

1. $\frac{x}{2x - 1}\ $where $x = 3$

\[\begin{array}{ll} \frac{x}{2x - 1} & \\ = \frac{3}{(2\left( 3 \right) - 1)} & \text{Substitute}^* \\ = \frac{3}{(6 - 1)} & \text{Parentheses: multiplication} \\ = \frac{3}{5} & \text{Parentheses: subtraction} \end{array} \]

*When we substituted, we also put the denominator in a set of parentheses. That helps us to keep track of the order of operations. We work on parentheses first, and within those parentheses, still follow the order of operations.

2. $x - (4 - x)$ where $x = - 1$

\[\begin{array}{ll} x - \left( 4 - x \right) & \\ = \left( - 1 \right) - \left( 4 - \left( - 1 \right) \right) & \text{Substitute} \\ = - 1 - \left( 4+ 1 \right) & \text{Change Subtraction} \\ = - 1 - \left( 5 \right) & \text{Parentheses} \\ = - 6 & \text{Subtraction} \end{array}\]

Alternatively, we could have chosen to simplify this expression, then substitute.

\[\begin{array}{ll} x - (4 - x) & \\ = x+\left( - 1 \right)(4 + \ - x) & \text{Change Subtraction}^{*} \\ = x + \ - 1 \cdot 4 +\left( - 1 \right)\left( - x \right) & \text{Distribute} \\ = x- 4 + x & \text{Multiply} \\ = 2x - 4 & \text{Combine like terms} \\ = 2\left( - 1 \right) - 4 & \text{Substitute} x = - 1 \\ = - 2 - 4 & \text{Multiplication} \\ = - 6 & \text{Subtraction} \end{array} \]
*When we changed the subtraction in front of the parentheses, we also wrote down the invisible 1 that is implied there. So we have $x - 1(4 - x)$, which can be rewritten as $x + ( - 1)(4 - x)$.

3. $\left( xy^{2} \right)^{3}$ where $x =\frac{1}{2}$ and $y = 2$

\[ \begin{array}{ll} \left( x y^{2} \right)^{3} & \\ = \left( \left( \frac{1}{2} \right)\left( 2 \right)^{2} \right)^{3} & \text{Substitute} \\ = \left( \left( \frac{1}{2} \right)\left( 4 \right) \right)^{3} & \text{Parentheses (exponents)} \\ = \left( \frac{4}{2} \right)^{3} & \text{Parentheses (multiplication)} \\ = 2^{3} & \text{Parentheses (division)} \\ = 8 & \text{Exponents} \end{array} \]

When we are determining whether a value of a variable is a valid solution to an equation, we want to evaluate each side of the equation for the given value of the variable and see if they are equal.

1. \[\begin{array}{rclr} x^{2} - 4 & = & 3x & \\ (0)^{2} - 4 & ?= & 3(0) & \text{Substitute } x = 0 \\ 0 - 4 & ?= & 0 & \text{Evaluate each side} \\ -4 & \neq & 0 & \end{array}\]

Let’s dig into what we’re doing here. First of all, while we’re checking our potential solution, we use a $? =$ instead of an $=$. We’re not claiming that the two sides are equal, we’re checking whether they are indeed equal.

Notice also that we’re working with each side independently, simplifying each expression using the order of operations.

Now, at the end, we have $- 4$ on the left and $0$ on the right, which are not equal. This means that the value we substituted in, $x = 0$, is not a solution for the equation, because it doesn’t cause the expression on the left to be equal to the expression on the right.

2. \[ \begin{array}{rclr} x^{2} - 4 & = & 3x & \\ (1)^{2} - 4 & ?= & 3(1) & \text{Substitute } x = 1 \\ 1 - 4 & ?= & 3 & \text{Evaluate each side} \\ -3 & \neq & 3 & \end{array} \]

When we substitute $x = 1$ into the equation, the left side evaluates to -3, but the right side evaluates to 3. Since $- 3 \neq 3$, $x = 1$ is not a solution to the equation.

3. \[ \begin{array}{rclr} x^{2} - 4 & = & 3x & \\ (-1)^{2} - 4 & ?= & 3(-1) & \text{Substitute } x = -1 \\ 1 - 4 & ?= & -3 & \text{Evaluate each side} \\ -3 & = & -3 & \end{array} \]

   This time, finally, when we substitute in -1 for $x$, we get a true statement at the end: $- 3 = - 3$. This means $x = - 1$ is a solution to the equation!

This time, before we worry about moving $x$ terms or undoing operations, it might be in our best interests to simplify first.

\[ \begin{array}{rclr} 4(x - 3) & = & 1 - 1(x + 2) & \text{Write invisible 1} \\ 4(x - 3) & = & 1 + \left( -1 \right)(x + 2) & \text{Change subtraction} \\ 4 \cdot x - 4 \cdot 3 & = & 1 + (-1) \cdot x + (-1) \cdot 2 & \text{Distribute} \\ 4x - 12 & = & 1 - x - 2 & \text{Simplify} \\ 4x - 12 & = & -x - 1 & \text{Combine like terms} \end{array} \]

In this first part of the process, we worked with each side individually, so we had to stick to do’s and don’ts of working with expressions (we were careful not to change the value of any expression). Now, we want to start moving things to the other side of the equal sign, so we’ll change both sides of the equation in the same way.

\[ \begin{array}{rclr} 4x - 12 & = & -x - 1 & \\ 4x - 12 + x & = & -x - 1 + x & \text{Add } x \text{ to both sides} \\ 4x + x - 12 & = & -x + x - 1 & \text{Rearrange} \\ 5x - 12 & = & 0x - 1 & \text{Combine like terms} \\ 5x - 12 & = & -1 & \text{Simplify} \\ 5x - 12 + 12 & = & -1 + 12 & \text{Add 12 to both sides} \\ 5x + 0 & = & 11 & \text{Simplify} \\ 5x & = & 11 & \\ \frac{5x}{5} & = & \frac{11}{5} & \text{Divide by 5 on both sides} \\ x & = & \frac{11}{5} & \text{Simplify} \end{array} \]

Let’s check that work by substituting $x = \frac{11}{5}$ into the original equation, $ 4(x - 3) = 1 - (x + 2)$.

\[ \begin{array}{rclr} 4\left( \frac{11}{5} - 3 \right) & ?= & 1 - \left( \frac{11}{5} + 2 \right) & \\ 4\left( \frac{11}{5} - \frac{3}{1} \cdot \frac{5}{5} \right) & ?= & 1 \cdot \frac{5}{5} - \left( \frac{11}{5} + \frac{2}{1} \cdot \frac{5}{5} \right) & \text{Make common denominator} \\ 4\left( \frac{11}{5} - \frac{15}{5} \right) & ?= & \frac{5}{5} - \left( \frac{11}{5} + \frac{10}{5} \right) & \text{Simplify} \\ 4\left( -\frac{4}{5} \right) & ?= & \frac{5}{5} - \frac{21}{5} & \text{Subtract} \\ \frac{4}{1} \cdot -\frac{4}{5} & ?= & \frac{5}{5} - \frac{21}{5} & \\ -\frac{16}{5} & = & -\frac{16}{5} & \end{array} \]

Looks like $x = \frac{11}{5}$ is a good solution to the equation!

1. Here we have $3 \cdot x = 0$. According to the ZPP, that’s only possible if either 3 equals 0 or x equals 0. Since $3 \neq 0$, the only solution is $x = 0$.

2. This time we have $3 \cdot (x - 2) = 0$. Again, $3 \neq 0$, so if we use the ZPP, we’ll only have $x - 2 = 0$. Then, we can finish isolating $x$ in that little equation.

\[\begin{array}{rcl} x - 2 & = & 0 \\ x - 2 + 2 & = & 0 + 2 \\ x & = & 2 \end{array}\]

So our solution to the equation is $x = 2$. Let’s double check that.

\[ \begin{array}{rclr} 3(x - 2) & = & 0 & \\ 3(2 - 2) & ?= & 0 & \text{Substitute } 2 \text{ for } x. \\ 3(0) & ?= & 0 & \text{Simplify} \\ 0 & = & 0 & \end{array} \]

We ended with a true statement ($0 = 0$), so it looks like $x = 2$ is a good solution.

3. This time we have a full $3x$ times our $x - 2$, so according to the ZPP, either $3x = 0$ or $x - 2 = 0$. That gives us two equations to finish solving for $x.$

We already found that when $3x = 0$, $x = 0$, and when $x - 2 = 0$, the solution is $x = 2$. That means our equation has two solutions, $x = 0$ and $x = 2.$

4. Now the things being multiplied are $(x + 3)$ and $(x - 2)$. According to the ZPP then, either $x + 3 = 0$ or $x - 2 = 0$. Isolate $x$ in each of these equations to get solutions of $x = - 3$ or $x = 2$.