This is the section where letters show up and a lot of students panic. Don't. A letter is just a number you haven't met yet. Almost every problem here comes down to two skills: cleaning up a messy expression, and undoing an equation one step at a time until $x$ is standing all by itself. The ISEE isn't trying to test whether you're a genius. It's testing whether you're careful with minus signs. So let's be careful with minus signs.
The practice is mixed into the lesson: each question sits right after the idea it needs, and the Mixed Practice at the bottom takes that hint away. Flip the switch above any question to type your answer instead. Numbers and fractions work (17/4), and so do expressions (5x + 2, 3(2x + 3)), equations for a letter (y = -2x + 7), inequalities (x > 2, -4 <= x <= 10, x < -3 or x > 5), and the words "no solution" or "infinitely many." A few versions need their answer choices, and they'll tell you so.
First, the vocabulary. You're going to see the words "expression" and "equation" all over the test, and they are not the same thing. But before either one, meet the letters.
A variable is a letter that stands for a number.
Mix variables and numbers together with some operations, and you've built an expression.
An expression is a combination of numbers, variables, and operation symbols that represents a value and does not include an equals sign.
So $3x-7$, $2(x+5)$, and $\dfrac{x}{3}+4$ are all expressions. No equals sign means there's nothing to solve. All you can do with an expression is simplify it or plug numbers into it. To simplify, you need to know what the pieces are called.
A term is one piece of an expression: a number, a variable, or numbers and variables multiplied together. Terms are separated by plus and minus signs, and the minus sign belongs to the term that follows. A term with no variable is a constant.
Read that last part again: the minus sign belongs to the term that follows. That one sentence is responsible for about half the wrong answers in this whole section. In $7x-3x+5$, the middle term isn't $3x$. It's $-3x$. The minus sign is glued to it and goes wherever it goes.
Every sign in front of a term is part of that term. Keep that in mind and the rest of this page gets a lot easier. The number stuck to the front of a term gets a name too.
The coefficient of a term is the number multiplied by its variable, sign included.
Here's all of that new vocabulary on one expression.
In $7x-3x+5$, name the terms, the constant, and the coefficients.
Split at the plus and minus signs, and keep each sign with the term after it: the terms are $7x$, $-3x$, and $5$.
The $5$ has no variable, so it's the constant. The coefficients are the numbers on the $x$-terms, signs included: $7$ and $-3$.
And when two terms have exactly the same letters, they're teammates.
Like terms have the same variable part (same variables to the same powers); only like terms can be combined.
Here's what that looks like on a messy expression, with each kind of term in its own color.
Simplify $1+2x+3x^{2}+4x-2x^{2}-3$.
$1$ $+$ $2x$ $+$ $3x^{2}$ $+$ $4x$ $+$ $-2x^{2}$ $+$ $-3$ $=$ $-2$ $+$ $6x$ $+$ $x^{2}$
Numbers: $1-3=-2$. $x$-terms: $2x+4x=6x$. $x^{2}$-terms: $3x^{2}-2x^{2}=x^{2}$. Each pile only combines with itself.
Think of it like sorting laundry. Socks go with socks, shirts go with shirts. $x$'s go with $x$'s, $x^{2}$'s go with $x^{2}$'s, and plain numbers go with plain numbers. You would never fold a sock into a shirt, and you never add $3x^{2}$ to $2x$. Try it yourself below.
Notice what the bins are really doing. Each one just adds up the numbers in front of its terms. That's the whole rule:
Combine Like Terms:
$$ax+bx=(a+b)x$$
Here it is on a real expression, minus signs and all.
Simplify: $3x-5+2x+7$.
The $x$-terms are $3x$ and $2x$. The numbers are $-5$ and $7$ (the minus sign goes with the $5$!).
$$3x+2x=5x \qquad -5+7=2$$
So the answer is $5x+2$. If you got $5x-12$, you let the minus sign on the $5$ spread to the $7$: that's $-5-7$ instead of $-5+7$.
Here's the same idea with tiles you can actually count. A positive tile and a negative tile of the same kind make a zero pair: together they're worth $0$, so they cancel. Whatever survives is the simplified expression.
Now test your eye. Some of these really are the same expression, and some only look like it.
Your turn. Sort the terms in your head, give every term its own sign, and add up each pile.
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Parentheses are the other thing standing between you and a simplified expression. The number out front has to multiply everything inside, not just the first thing it bumps into.
The distributive property says $a(b+c)=ab+ac$: the number out front multiplies every term inside. Run it backward, $ab+ac=a(b+c)$, and it's called factoring.
Same rule, two directions. Distributing opens the parentheses up; factoring packs them back in. Why does the number out front hit both terms? Picture it as the height of a rectangle. The area is both pieces, not just the first one.
Now a real one, with two sets of parentheses to open.
Which is equivalent to $3(x-4)+2(x+1)$?
Distribute each group separately:
$$3(x-4)=3x-12 \qquad 2(x+1)=2x+2$$
Now combine like terms: $3x+2x=5x$ and $-12+2=-10$. The answer is $5x-10$.
The ISEE likes to trick students here by putting $5x-14$ in the choices. That's a sign slip: $-12-2$ instead of $-12+2$. And if you forget to multiply the $3$ onto the $-4$, you get $3x-4+2x+2=5x-2$, which is just as wrong. The number outside hits both terms.
Now the sneaky one. A minus sign in front of parentheses is really a $-1$ being distributed, and it flips the sign of every term inside.
Simplify: $(5x-2)-(3x-7)$.
The minus sign hits both terms in the second group: $-(3x-7)=-3x+7$. So
$$5x-2-3x+7=2x+5$$
The trap answer is $2x-9$. That's what happens when the minus sign reaches the $3x$ but not the $-7$. Minus a negative is a plus!
That one is easy to understand and easy to forget in the middle of a test. So here's a habit.
When you see a minus sign in front of parentheses, rewrite it as $-1(\ldots)$ before you do anything else. It feels slow. It's not nearly as slow as getting the problem wrong.
Watch the number out front hit every term, and then watch what happens when it doesn't.
Put all of it together and you get the full warm-up: distributing, minus signs, and like terms in one problem.
Simplify completely: $3x-5+2x(4-3x)+7x^{2}-6(x-1)$.
Distribute first. Careful: $2x$ times $-3x$ is $-6x^{2}$, and $-6$ times $-1$ is $+6$.
$$3x-5+8x-6x^{2}+7x^{2}-6x+6$$
Now sort into bins. $x^{2}$-terms: $-6x^{2}+7x^{2}=x^{2}$. $x$-terms: $3x+8x-6x=5x$. Numbers: $-5+6=1$.
$$x^{2}+5x+1$$
Your turn: distribute first, then combine like terms.
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Factoring runs the same rule backward. Find the biggest number that divides every term, and pull it out front.
Factor completely: $8x-20$.
Both $8$ and $20$ are divisible by $4$, and nothing bigger goes into both. Divide each term by $4$: $8x \div 4 = 2x$ and $-20 \div 4 = -5$. So
$$8x-20=4(2x-5)$$
Check by distributing back: $4(2x)-4(5)=8x-20$. If you wrote $2(4x-10)$, that's equal, but it isn't completely factored: $4x$ and $10$ still share a $2$.
Your turn. Pull out the greatest common factor, and check by distributing back.
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The last thing you can do with an expression is plug numbers in. This is where negative numbers love to cause trouble.
Remember $(-3)^{2}=9$ but $-3^{2}=-9$ from Section 2? Plugging in a negative is exactly when that matters, so the parentheses go on first.
If $a=-2$ and $b=3$, evaluate $2a^{2}-3b$.
Put the negative number in parentheses before you square it: $(-2)^{2}=4$. Then
$$2(4)-3(3)=8-9=-1$$
Your gut might say $(-2)^{2}=-4$. Your gut is wrong. A negative times a negative is positive. That mistake gives $-8-9=-17$. And if you forget the $2$ out front, you get $4-9=-5$, another trap choice.
Your turn. Parentheses on first, then the exponent, then everything else.
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Now we add an equals sign, and suddenly there's something to find.
An equation says two expressions are equal, so it has an equals sign. To solve an equation, find every value of the variable that makes it true. Those values are the solution.
The best way to picture an equation is a balance scale. The left side and the right side weigh exactly the same. You're allowed to do anything you want to the scale (add weight, take weight off, cut everything in half) as long as you do it to both sides, because then it stays balanced. Your mission is to keep doing that until one pan holds a single $x$ and nothing else. Whatever's on the other pan is the answer.
Once you've solved one, flip to "Left Side Only" and watch what happens.
"Whatever you do to one side, you must do to the other." That's the only rule. The order that works best: get all the $x$'s on one side, then get the numbers off that side, then divide.
Solve: $5x-7=3x+9$.
Take $3x$ off both sides: $2x-7=9$.
Add $7$ to both sides: $2x=16$.
Divide both sides by $2$: $x=8$.
One version of this looks scarier than it is.
If $(8.05 + 1.95)t = t$, what is $t$? The decimals are a costume: $8.05 + 1.95 = 10$, so this is $10t = t$. Subtract $t$: $9t = 0$, so $t = 0$. The trap is dividing both sides by $t$ to get $10 = 1$. You can't divide by a letter that might be $0$, and here it is $0$: the equation's only solution.
Your turn: $x$'s to one side, numbers to the other, then divide.
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Why undo things in that order? Think of $x$ as a present that got wrapped: the last layer on is the first layer off.
So in $4x+7=31$ the $+\,7$ comes off first, and the $\times\, 4$ comes off last. Try a few two-step equations.
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If there are parentheses, distribute first. Then it's the same three steps.
Solve: $3(x-4)=2x+5$.
Distribute: $3x-12=2x+5$. Take $2x$ off both sides: $x-12=5$. Add $12$: $x=17$.
If you got $9$, you wrote $3x-4$ instead of $3x-12$. Hit both terms!
And here's our old friend, the minus sign in front of parentheses.
Solve: $3x-(2-4x)=19$.
The minus flips both signs inside: $3x-2+4x=19$. Combine: $7x-2=19$. Add $2$: $7x=21$. So $x=3$.
If you only flipped the $2$, you'd get $3x-2-4x=19$, which leads to $x=-21$. Not even close.
Your turn. Clear the parentheses first, minus signs included.
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Word problems are just equations wearing a disguise. Translate one phrase at a time.
"Twice a number decreased by 7 equals 11." Find the number.
"Twice a number" is $2x$. "Decreased by 7" means subtract $7$. "Equals 11" is $=11$.
$$2x-7=11, \quad\text{so}\quad 2x=18, \quad\text{so}\quad x=9$$
One phrase in particular trips everyone up, because it reads backwards.
Watch out for "less than." "7 less than twice a number" means $2x-7$, not $7-2x$. It's a favorite on the ISEE because it reads backwards. If you start with $10$ and go $7$ less than it, you get $10-7$. The $7$ comes off the end.
Translating is a skill all by itself, so practice it on its own first. Build each sentence's equation from the tiles, one phrase at a time.
And here's the order trap by itself. Each phrase means either $2n-7$ or $7-2n$. Which one?
Now translate and solve.
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Fractions look scary, but there's one move that makes them vanish: multiply every single term on both sides by a number that all the denominators go into.
Here's that one written out, with a check at the end.
Solve: $\dfrac{x}{3}+\dfrac{x}{4}=14$.
Both $3$ and $4$ go into $12$, so multiply every term by $12$:
$$12\cdot\frac{x}{3}+12\cdot\frac{x}{4}=12\cdot 14, \quad\text{so}\quad 4x+3x=168$$
So $7x=168$ and $x=24$. Quick check: $\frac{24}{3}+\frac{24}{4}=8+6=14$. ✓
Same trick when there's a fraction on each side.
Solve for $x$ when $\dfrac{3x-5}{3}=\dfrac{2x+7}{6}$.
Multiply both sides by $6$. The left side becomes $2(3x-5)$ and the right side becomes $2x+7$:
$$6x-10=2x+7, \quad\text{so}\quad 4x=17, \quad\text{so}\quad x=\frac{17}{4}$$
Yes, a fraction is a perfectly good answer. Don't panic and assume you made a mistake.
Notice that the $2$ had to hit the whole numerator. That's worth a reminder.
When you multiply to clear the fractions, put parentheses around each numerator first. The $2$ in $2(3x-5)$ has to hit the $-5$ too. Forget that, and you're back to the distributive trap.
Your turn: clear the fractions, then solve.
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Sometimes an equation has more than one letter in it, and the question only wants one of them.
A literal equation is an equation with more than one letter. Solving it for one letter means getting that letter alone.
Same balance-scale moves as always. The other letters just come along for the ride.
If $x - y = 3$, which expression is equal to $y$?
Subtract $x$ from both sides: $-y = 3 - x$. Multiply both sides by $-1$: $y = x - 3$.
That last step, multiplying by $-1$, is exactly where students slip.
Multiplying by $-1$ flips every sign on that side, not just the first one. The ISEE puts $x + 3$, $-x + 3$, and $-x - 3$ in the choices to catch half-flipped answers. Quick check: pick $x = 5$. Then $y$ should be $2$, and only $x - 3$ gives $2$.
Formulas are literal equations too, and they work exactly the same way.
The perimeter of a rectangle is $P = 2l + 2w$. Solve for $l$.
Subtract $2w$ from both sides: $P - 2w = 2l$.
Divide both sides by $2$, and that means every term on both sides: $l = \dfrac{P - 2w}{2}$.
The trap is $l = \dfrac{P}{2} - 2w$, where only the $P$ got divided.
Try it with moves of your own. Each one happens to both sides.
Your turn. Get the letter they ask for alone.
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Other times there are too many letters to solve for each one, and you don't need to.
If $3m + 3n + p = 30$ and $m + n = 8$, what is the value of $p$?
You can't find $m$ and $n$ separately, but you don't have to. Group them: $3m + 3n = 3(m + n) = 3(8) = 24.$
So $24 + p = 30$, and $p = 6$.
Here's that move in slow motion: spot the chunk, then swap in its value.
So the habit is simple.
When a question gives you a chunk like $m + n$, look for that same chunk hiding in the other equation. If you start solving for $m$ by itself, you're doing it the hard way (and usually the impossible way).
Your turn: find the chunk, then use it.
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One more trick: if the same chunk shows up on both sides, pull it out and cancel it.
Pulling a common chunk out of every term is called factoring, and Section 9 turns it into a whole toolkit.
If $3x - 3 = xy - y$ and $x \neq 1$, what is the value of $y$?
Factor both sides: $3(x - 1) = y(x - 1)$.
Since $x \neq 1$, the chunk $(x - 1)$ isn't zero, so divide both sides by it: $y = 3$.
That "$x \neq 1$" is doing real work: you can't divide by a chunk that might be zero. Your turn.
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Most equations have exactly one answer. But sometimes, while you're solving, every $x$ disappears. When that happens, look at what's left over.
An identity is an equation that is true for every value of the variable. It has infinitely many solutions.
And the exact opposite can happen too.
An equation has no solution when no value of the variable makes it true.
Here's one of each.
How many solutions do $2(x+3)=2x+6$ and $x=x+1$ have?
Distribute the first: $2x+6=2x+6$. Both sides are the same expression, so every $x$ works. It's an identity, with infinitely many solutions.
In the second, take $x$ off both sides: $0=1$. That's never true, so there is no solution.
Load both of these into the Balance Scale above. With $2(x+3)=2x+6$, the two pans are identical, so they balance no matter what $x$ is. With $x=x+1$, the right pan is always heavier by exactly $1$, so the scale can never balance.
Here's another way to see all three cases. Make each side of the equation a bar, and slide $x$ around looking for a spot where the bars match.
So here's the rule of thumb:
Here's the rule in action on a typical ISEE question.
Which equation has infinitely many solutions?
Distribute $2(x+3)$ and you get $2x+6$, exactly the other side. That's an identity, so it has infinitely many solutions.
The first and last have matching $x$'s but different numbers, so they have no solution. And $4(x-2)=3x-8$ becomes $4x-8=3x-8$, which has different $x$-coefficients. It has exactly one solution ($x=0$).
Fractions don't change anything: clear them first, then look at what's left.
Okay, but what if a no-solution line like $6=-1$ did count? Add $1$: $7=0$. Times $4$: $28=0$. Add $72$: $100=72$. So my $72$ on the last quiz was secretly a $100$... Fine. No solution.
Solve: $\dfrac{x+3}{5}=\dfrac{2x-1}{10}$.
Multiply both sides by $10$: $2(x+3)=2x-1$, so $2x+6=2x-1$.
Take $2x$ off both sides: $6=-1$. That's never true, so there is no solution.
If you forgot to multiply the $3$ by $2$, you'd get $2x+3=2x-1$. Still no solution, so you'd get lucky this time. Don't count on luck.
Your turn to sort them. Simplify each one in your head first.
Now on your own. Some of these ask which equation, and some ask you to solve one.
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An equation says two things are equal. An inequality says one is bigger.
An inequality states that one expression is less than ($<$), at most ($\le$), greater than ($>$), or at least ($\ge$) another.
The good news: you solve inequalities exactly the way you solve equations. Same steps, same order. The bad news: there's one rule that equations don't have, and the ISEE tests it constantly.
Inequality Rule: multiplying or dividing both sides by a negative flips $<$ to $>$ and $\le$ to $\ge$ (and vice versa).
Why? Because negatives reverse the order of numbers. $2<5$, but $-2>-5$ (negative two is warmer than negative five). Multiply by a negative and the bigger side becomes the smaller side, so the sign has to turn around. Slide the multiplier and watch it happen.
Okay, negatives flip it. What about multiplying both sides by $0$? $2<5$ becomes $0<0$. Which is false. So zero doesn't flip it, it just breaks it. Fine. I'll stick to the rule.
Here's the flip in a real problem.
Solve the inequality: $-3x+9<0$.
Subtract $9$: $-3x<-9$.
Divide by $-3$, and flip the sign: $x>3$.
The trap answer is $x<3$, which is what you get when you divide and forget to flip. Quick check: $x=4$ gives $-12+9=-3<0$. ✓
Not every minus sign means a flip, though. Only multiplying or dividing by a negative does.
Solve: $4-2(3-x)\ge x+1$.
Distribute the $-2$ to both terms: $4-6+2x\ge x+1$, so $2x-2\ge x+1$.
Take $x$ off both sides: $x-2\ge 1$. Add $2$: $x\ge 3$.
No negative division here, so no flip. Don't flip just because you see a minus sign somewhere!
And fractions? Clear them first. Multiplying by a positive number never flips anything.
Solve the inequality: $\dfrac{x-1}{2}\ge\dfrac{x+3}{4}$.
Multiply both sides by $4$ (positive, so no flip): $2(x-1)\ge x+3$, so $2x-2\ge x+3$.
Take $x$ off and add $2$: $x\ge 5$.
Your turn. Watch for the flip, and only flip when it's earned.
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Here's the best safety net for every one of those.
Not sure if you flipped correctly? Plug in a test number. Pick something that your answer says should work, stick it into the original inequality, and see if it's actually true. It takes ten seconds and catches almost every flip mistake.
Try it: predict which side gets shaded, then let a test point check you.
Testing a single number is a question on its own, too. Which of these numbers actually works?
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Sometimes $x$ is trapped between two numbers at once.
A compound inequality puts an expression between two bounds at once.
Solve it the same way, but every move has to happen to all three parts.
Which values of $x$ satisfy $41 \le 2x - 1 \le 51$?
Add $1$ to all three parts: $42 \le 2x \le 52$. Divide all three by $2$: $21 \le x \le 26$.
On a number line, that's a closed dot at $21$, a closed dot at $26$, and everything between them shaded. If a question asks how many whole numbers work, count both ends: $26-21+1=6$ of them.
Here's that squeeze on a number line: adding slides the range, dividing shrinks it.
Dividing by a negative works too, with one twist.
If you divide a compound inequality by a negative, both signs flip, so the two ends swap places. $-6 \le -2x \le 4$ becomes $3 \ge x \ge -2$, which reads better as $-2 \le x \le 3$.
The test asks this both ways: solve and pick the matching number line, or read a number line and pick the matching inequality. Either way, check the dots (open or closed) and whether the shading is between the ends or outside them.
Your turn. Every move, all three parts.
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Absolute value has a reputation for being confusing, but it really comes down to one word: distance.
The absolute value $|x|$ is the distance of $x$ from $0$ on the number line, so it's never negative.
So $|5|=5$ and $|-5|=5$: both numbers are $5$ steps from $0$. When there's arithmetic inside the bars, do it first, then take the distance.
Evaluate $|x-4|+|x|$ when $x=-1$.
Inside first: $x-4=-1-4=-5$, and $x=-1$. Then the distances: $|-5|=5$ and $|-1|=1$. So the value is $5+1=6$.
If you got $-6$, you dropped the bars. They turn $-5$ into $5$, never the other way around.
Your turn. Inside the bars first, then the distance.
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Distance between two numbers on a line is $|a-b|$. In Section 5 the same idea grows up into the distance between two points on a graph.
So $|x-3|$ is the distance from $x$ to $3$. Saying $|x-3|=7$ is saying "$x$ is exactly $7$ away from $3$," and there are two numbers like that: one $7$ to the right, one $7$ to the left. That's why absolute value equations usually have two answers.
Absolute Value: for $k \ge 0$,
$$|x-a|=k \quad\text{means}\quad x=a+k \ \text{ or } \ x=a-k$$
A distance is never negative, so when $k$ is negative (like $|x-a|=-3$) there's no solution at all.
Watch it happen on a number line, plus sign inside and all.
Here's the same equation as a test question.
Let $|x+2|=5$. What is the sum of all solutions?
Rewrite $x+2$ as $x-(-2)$: this says $x$ is $5$ away from $-2$. So $x=-2+5=3$ or $x=-2-5=-7$.
The sum is $3+(-7)=-4$. The trap answer is $4$, from treating the center as $+2$ instead of $-2$. A plus sign inside means the center is negative.
Your turn. Find the center first.
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The same distance idea handles inequalities. "At most $7$ away" is everything squeezed in the middle. "At least $7$ away" is everything out past the edges, heading off in both directions.
Absolute Value Inequalities:
$$|x-a|\le k \quad\text{means}\quad a-k\le x\le a+k$$
$$|x-a|\ge k \quad\text{means}\quad x\le a-k\ \text{ or }\ x\ge a+k$$
Play with it below. The caption writes out the inequality for you, and the little test pointer underneath checks any number you like.
Here are the two shapes on paper. First, "at most": the middle.
Solve: $|x-3|\le 7$.
$x$ is at most $7$ away from $3$, so it's squeezed between $3-7$ and $3+7$:
$$-4\le x\le 10$$
That's every whole number from $-4$ to $10$, which is $10-(-4)+1=15$ of them.
Then "more than": the two pieces out past the edges.
Solve: $|x-1|>4$.
$x$ is more than $4$ away from $1$, so it's out past one of the edges: below $1-4=-3$ or above $1+4=5$.
$$x<-3 \quad\text{or}\quad x>5$$
That's two separate pieces, so the answer needs the word "or." Writing $-3 < x < 5$ is the trap: that's the middle, the numbers within $4$ of $1$.
Your turn. Middle or edges?
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Two answers is the usual case for an equation, but not the only one. Predict before you peek.
When there's an $x$ on both sides, you can't just use the distance picture. Split it into two plain equations instead: the inside equals the right side, or the inside equals the negative of the right side.
Solve: $|2x-5|=x+4$.
Case 1: $2x-5=x+4$, so $x=9$.
Case 2: $2x-5=-(x+4)=-x-4$, so $3x=1$ and $x=\frac{1}{3}$.
Now check both, because an absolute value can never equal a negative number. At $x=9$: $|13|=13$ ✓. At $x=\frac{1}{3}$: $\left|-\frac{13}{3}\right|=\frac{13}{3}$ ✓. Both work.
Both of those survived the check. They don't always.
And one more habit for Case 2.
Put parentheses around the right side before you make it negative in Case 2. $-(x+4)$ is $-x-4$, not $-x+4$. Sound familiar? It's the same minus-sign trap from the top of this page, just wearing a different outfit.
Your turn to be the judge: keep each candidate, or throw it out?
Now solve some all the way through: two cases, then the check.
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The ISEE loves quantitative comparison, and with algebra it has a favorite trick: give you an $x$ with no value and hope you panic and pick (D), "cannot be determined." Don't. Simplify first. Lots of the time the $x$'s cancel and the answer is settled no matter what $x$ is.
Distribute: Column A is $2x+10$. Both columns have $2x$, and $10>8$, so Column A is greater, whatever $x$ is. The $x$ was a bluff.
Don't believe it? Watch the two columns while $x$ moves around.
When they do give you a value, plug it in carefully, especially inside absolute value bars.
Let $x=-1$.
Column A: $|-1-4|=|-5|=5$. Column B: $|4-(-1)|=|5|=5$. They're equal.
And they always will be: the distance from $x$ to $4$ is the same as the distance from $4$ to $x$. You didn't even need the value.
And sometimes each column is its own little equation to solve.
Solve each one. Column A: $4x=16$, so $x=4$. Column B: $3x=18$, so $x=6$. Column B is greater.
One last rule of thumb, for choice (D).
Only pick (D) when you can actually find two allowed values of $x$ that give different winners. If you can't break it, it's probably not (D).
Every question so far came right after its idea, which is a hint the real test never gives. Down here the hint is gone. First, the quantitative comparison: two columns, four fixed choices, and any idea from this section fair game.
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Next, the mixed set. Every question type on this page except the quantitative comparison goes into one hat, and each click draws a new one at random. Flip the switch to type a number answer; answers that are expressions or inequalities stay multiple choice down here.
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Alright, time for something with a little heat. These are absolute value inequalities with an $x$ on both sides. The trick is to trap the inside between the right side and its negative, like $-(x+3)\le 2x-6\le x+3$, then solve each half separately. For "greater than," split it instead: the inside is at least the right side, or at most its negative. Either way, watch your signs when you make the right side negative.
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