Foundations of Algebra

Solution. undo the "+ 7" by subtracting 7 from both sides (whatever we do to one side we must do to the other): $x+7-7=12-7$. Step 2 — simplify each side: $x=5$. Sanity check — put the answer back into the original: $5+7=12$. Correct.

The symbol $x$ stood for the specific number $5$, but the method would work for "a number plus 7 equals anything."□

Solution. choose values for the variables, say $l=5$ and $w=3$. Step 2 — substitute them in: $P=2(5)+2(3)$. Step 3 — evaluate, multiplication before addition: $P=10+6=16$.

The very same formula gives $P=2(7)+2(2)=18$ for a $7\times 2$ rectangle — one formula, infinitely many rectangles.□

Solution. assign a variable: let the number be $x$. Step 2 — translate the sentence into symbols. "Doubled" is $2x$; "4 subtracted" is $2x-4$; "the result is 10" is $=10$. So $2x-4=10$. Step 3 — undo the subtraction: add 4 to both sides, $2x=14$. Step 4 — undo the doubling: divide both sides by 2 (allowed because $2\ne 0$), $x=7$. Sanity check — reread the story with $x=7$: double it to get $14$, subtract 4 to get $10$. Correct.

Notice the "translate, then solve, then check" pattern — it is the whole job of algebra in miniature.□

Solution. Here $x$ is the variable — the symbol for the unknown number. $4$ and $10$ are constants — fixed, known numbers. $+$ and $=$ are the operator and the relation that tie them together.

Solving gives $x=6$ (subtract 4 from both sides). Before solving, $x$ was a box with a question mark; after solving, we know the box held $6$.□

Solution. substitute $b=8$ and $h=5$: $A=\frac{1}{2}\times 8\times 5$. Step 2 — evaluate left to right: $\frac{1}{2}\times 8=4$, then $4\times 5=20$.

So $A=20$ square units. Change the inputs to $b=10,h=3$ and the same machine returns $A=\frac{1}{2}\times 10\times 3=15$ — the formula is a reusable machine, and $b$ and $h$ are the slots you drop numbers into.□

Solution. In $3x^2-5x+2=0$, matching against the template gives $a=3$ (coefficient of $x^2$), $b=-5$ (coefficient of $x$, sign included), and $c=2$ (the constant term). The $-5$ must carry its minus sign: the coefficient is $-5$, not $5$.

Reading coefficients by their conventional roles lets you drop them straight into a formula like the quadratic formula without re-deriving what each letter means.□

Solution. It has a variable term $4x$ (coefficient $4$ times the variable $x$) and a constant term $-9$, joined by subtraction. In words it says "take four times some number and subtract nine." It is a phrase, not a claim: it does not say it equals anything. Only when we write $4x-9=7$ do we have an equation to solve.

Its value depends on $x$: at $x=5$ it is $4(5)-9=11$; at $x=0$ it is $-9$. Different input, different value — but always just a value, never true or false on its own.□

Solution. substitute, using parentheses around the value to protect signs: $(4{)}^2-3(4)+2$. Step 2 — exponents first: $(4{)}^2=16$, giving $16-3(4)+2$. Step 3 — multiplication next: $3(4)=12$, giving $16-12+2$. Step 4 — addition/subtraction left to right: $16-12=4$, then $4+2=6$.

So the value is $6$. At $x=0$: $(0{)}^2-3(0)+2=0-0+2=2$. The parentheses matter most with negatives: at $x=-2$, $(-2{)}^2-3(-2)+2=4+6+2=12$ — squaring the $-2$ gives $+4$, and subtracting $3(-2)=-6$ adds 6.□

Solution. identify like terms: $5x$ and $-3x$ are both "$x$ terms"; $2$ and $7$ are both constants. Step 2 — group them: $(5x-3x)+(2+7)$. Step 3 — combine each group: $2x+9$.

The result $2x+9$ equals the original for every value of $x$ (check at $x=1$: original $5+2-3+7=11$; simplified $2+9=11$).□

Solution. the distributive property says $a(b+c)=ab+ac$, so multiply the $2$ into each term inside: $2\cdot x+2\cdot 4$. Step 2 — simplify: $2x+8$.

A common trap is to write $2x+4$, forgetting to multiply the $4$. Distribution touches every term inside the parentheses.□

Solution. The left-hand side is the expression $x+3$; the right-hand side is the expression $10$; the $=$ claims they have the same value. It is a conditional equation: it is true only for the one value $x=7$ (since $7+3=10$) and false for every other $x$ — e.g. at $x=2$ we get $5=10$, false.

Contrast with an identity like $2x=x+x$, which is true for all $x$, and a contradiction like $x=x+1$, which is true for no $x$. Solving means finding exactly which of these three cases you are in, and, when conditional, which value(s) work.□

Solution. undo the "$-5$ " by adding 5 to both sides: $3x-5+5=16+5$, so $3x=21$. Step 2 — undo the "$\times 3$ " by dividing both sides by 3 (legal because $3\ne 0$): $\frac{3x}{3}=\frac{21}{3}$, so $x=7$. Sanity check — substitute back into the original: $3(7)-5=21-5=16$. Matches the right-hand side.

Undo operations in reverse of how they were applied: the $x$ was multiplied by 3 then had 5 subtracted, so we add back the 5 first and divide by 3 last.□

Solution. substitute $x=7$ into the original equation: $2(7-3)$. Step 2 — evaluate: $2(4)=8$. Step 3 — compare with the right-hand side: $8=8$. The candidate checks out, so $x=7$ is a genuine solution.

Checking is not busywork: had a slip given us $x=5$, the check $2(5-3)=4\ne 8$ would flag it immediately.□

Solution. $3<7$: is 3 less than 7? Yes — true. $-2>1$: is $-2$ greater than 1? On the number line $-2$ sits to the left of 1, so no — false. $x\le 5$: this involves a variable, so instead of true/false it has a solution set — every number that is 5 or smaller. That includes $5$ itself (the "$=$ " part), $0$, $-3$, $4.9$, and so on, but not $6$.

Unlike an equation, an inequality with a variable usually has infinitely many solutions — a whole range, not a single value.□

Solution. subtract 1 from both sides (adding/subtracting never flips): $-2x\ge 6$. Step 2 — divide both sides by $-2$. Because the divisor is negative, flip $\ge$ to $\le$: $x\le -3$. Sanity check — pick a value in the claimed solution, say $x=-4$: $-2(-4)+1=8+1=9\ge 7$. Works. Now test a value outside it, $x=0$: $-2(0)+1=1\ge 7$ is false, as it should be.

Contrast with a positive divisor: solving $3x\ge 12$ gives $x\ge 4$ with no flip — the direction only reverses for a negative multiplier or divisor.□

Solution. mark the boundary at $2$. Because the inequality is strict ($>$, not $\ge$), draw an open circle at $2$: the point $2$ itself is not a solution ($2>2$ is false). Step 2 — decide the direction: solutions are numbers greater than 2, which lie to the right, so shade rightward and draw an arrow toward $+\infty$.

If instead we had $x\ge 2$, the only change would be a closed (filled) circle at $2$, because now $2$ satisfies $2\ge 2$ and belongs to the solution set.□

Solution. whatever we do, do it to all three parts to keep them lined up. To isolate $x$, subtract 2 everywhere: $-1-2\le x+2-2<5-2$. Step 2 — simplify each part: $-3\le x<3$. Sanity check — the endpoints tell the story: $x=-3$ is included (since $-1\le -3+2=-1$ holds with equality), but $x=3$ is excluded (since $3+2=5$ is not $<5$). A middle value like $x=0$ gives $-1\le 2<5$. True.

The solution is every $x$ from $-3$ inclusive up to $3$ exclusive.□

Solution.

1. Multiplication first (left to right): $4\times 2=8$.
2. Then addition and subtraction left to right: $3+8=11$, then $11-1=10$.

Solution. No agreed order — someone reading strictly left to right computes $3+4=7$, then $7\times 2=14$. PEMDAS order — multiplication outranks addition, so $4\times 2=8$ first, then $3+8=11$.

Two readers, two different numbers ($14$ vs. $11$) from the same symbols. The order-of-operations convention exists precisely to rule out one of them: the agreed answer is $11$. (The viral $6÷2(1+2)$ debates arise not because PEMDAS fails, but because the notation $2(1+2)$ is written ambiguously — the cure is to add parentheses, e.g. $6÷[2(1+2)]$ or $(6÷2)(1+2)$, so there is nothing left to argue about.)□

Solution.

1. Innermost parentheses: $2+3=5$.
2. Exponent: $5^2=25$.
3. Multiplication: $25\times 4=100$.

Solution.

1. Division and multiplication left to right: $6÷2=3$, $3\times 4=12$.
2. Addition and subtraction left to right: $8-3=5$, $5+12=17$.