Equations and Inequalities

Equations with Brackets #

Expand first, then the usual moves.

An equation with a bracket is expanded first, then solved with the usual moves

Before any balance move, multiply the bracket out.

Expand first — the 3 multiplies both parts inside, just as with brackets alone.

Now it is a shape you already solve: take the 6 off, divide by 3 — x is 3.

The usual slip: only the x gets multiplied, and the 2 slides through untouched.

Treating a Bracket as a Single Quantity #

Divide first and the bracket answers itself.

A bracket can be solved for whole, before anything inside it is touched

Expanding works and always will: 3x + 3 = 12, then 3x = 9, so x = 3.

Read it as three lots of x + 1. Divide by 3 and the bracket stands alone.

When the question wants 2a − b, one move hands it over. Neither letter is needed.

4x − 10 is 2x − 5 doubled, so it is twice 7 — and x never had to be found.

Equations with the Unknown on Both Sides #

Clear the smaller letter term off each side.

A letter on both sides is cleared by taking the smaller letter term off each side

Letters sit in both pans here, so no single move leaves x on its own.

Take 2x off both pans. The scales stay level, and one side has no letter left.

Written down, that move leaves 3x + 3 = 12 — the shape you already solve.

Now finish the old way: take the 3 off, divide by 3, and x comes out as 3.

How Many Solutions an Equation Has #

None, exactly one, or every number.

When the letters cancel away, a false line means no solution and a true line means every number

Each side is a line. 2x + 1 and 2x + 5 climb alike, so they never meet.

The algebra agrees: take 2x off both sides and 1 = 5 is left. No solution.

Here the two sides draw the same line. Every point on it is a solution.

4 = 4 is true whatever x is, so every number solves it. Infinitely many.

Different gradients cross once, and that crossing is the only solution.

An x survives the clearing and pins that crossing down: x = 2.

Choosing a Coefficient to Fix the Solution Count #

Match the x terms and the constants decide.

Matching the x terms is what decides whether the constants then agree or contradict

Whatever a turns out to be, the equation gathers down to (a − 5)x = 4.

Choose a = 5 and the x term dies. Zero lots of x never reach 4: none.

Any other a leaves a live x term, and a live x term pins x to one value.

Every number needs both: x terms that match, and constants that agree.

Equations with Fractions #

Multiply through and they vanish.

Multiplying every term by the denominator clears the fractions away

$\frac{x}{3}=4$x/3 = 4: a third of x is 4. The fraction is in the way.

Multiply both sides by 3 and the equation reads x = 12.

Changing the Subject of a Formula #

Rearrange until the letter you want stands alone.

Making a letter the subject means undoing what is done to it, both sides at once

The subject is the letter left alone. To free x, take the c off both sides.

Then divide both sides by m. The whole side is divided, never just one part.

Turn it round: $x=\frac{\left(y-c\right)}{m}$x = (y − c)/m. Same formula, and now x is the one standing alone.

Want m alone? Same start, but divide by x — the letter you pick sets the last move.

Making a Twice-Appearing Letter the Subject #

Gather, factor, then share.

A letter appearing twice is gathered and factored out before the final divide

There is an x term on each side, so no single division can isolate x.

Start by bringing every x term to one side. The c stays on the other side.

Now factor: ax − bx is x times (a − b), so the two x terms become one.

Divide both sides by the whole bracket (a − b), and x is the subject: $x=\frac{c}{\left(a-b\right)}$x = c/(a − b).

Forming Equations #

Name the unknown and the sentence solves.

Naming the unknown as a letter turns a word problem into an equation

Call the smaller number n and the larger one is n + 3.

Together they make 17, so n + (n + 3) = 17.

Writing Formulas from Words #

The story in five symbols.

A formula writes the words as letters, and a counted case checks it

Rides cost $p each and entry is $3: every extra ride adds one more p to the bill.

C = np + 3: the $3 entry is paid once, and $p is charged for each of the n rides.

The same structure fits another story: $s a week on top of $10 saved gives T = ws + 10.

Check with a case you can count: 2 rides at $5 is $10, plus the $3 entry, so C = 13.

Simultaneous by Substitution #

Put one equation into the other.

Substitution solves simultaneous equations by putting one into the other

Both equations hold at the crossing point, and one of them already says what y equals.

Substitute that y into the other equation: 2x + x collects to 3x, so x = 3 and y = 4.

Simultaneous by Elimination #

Add or subtract to lose a letter.

Elimination solves simultaneous equations by adding or subtracting them

One line is x + y = 10 and the other is x − y = 4; both hold where they cross.

Add them: +y and −y cancel, so 2x = 14 and x = 7. Then 7 + y = 10 gives y = 3.

Solving Simultaneous Equations by Scaling #

Scale until a letter lines up.

Scaling one equation lines up a letter so elimination can cancel it

Adding or subtracting these cancels nothing: no letter has the same coefficient in both.

Multiply the second equation by 2, and its 2y matches the 2y in the first equation.

Subtract them and the matching 2y terms cancel, leaving x = 2.

Substitute x = 2 back into x + y = 5 to get y = 3.

Writing a Pair of Equations from a Word Problem #

Name both unknowns, then say each fact once.

Two unknowns need two facts, and the answer is read back in the words it came from

Name both unknowns before anything else: a adult tickets and c child tickets.

Two facts give two equations: 9 tickets altogether, and a bill of 57.

Scale and subtract as before: a = 4 and c = 5.

A 4 on its own is not the answer: it is 4 adult tickets, and 32 plus 25 makes 57.

How Many Solutions a Pair of Equations Has #

Scale one and see whether it becomes the other.

Scaling one equation to match the other decides between none and infinitely many

These two lines are equally steep, so they run alongside and never meet.

Doubling matches the letters and not the number: 10 against 14. No solution.

Change that 14 to 10 and the two equations draw one single line.

Now doubling reproduces it whole, so every point on that line solves both.

No scaling turns 2x into 3x while keeping y, so these cross once, at (3, −1).

Choosing a Coefficient in a Pair of Equations #

Match the letters first, then look at the numbers.

Choosing the coefficient that makes one equation a multiple of the other decides how many solutions there are

Doubling the first equation gives 4x + 2y = 10, whatever k turns out to be.

At k = 4 the letters match and the numbers clash, so nothing solves both.

Equal gradients and different heights: the two lines run alongside and never meet.

Change that 12 to 10 and k = 4 reproduces the first equation whole.

Any other k tilts the second line differently, so the pair crosses once.

Solving for x + y Without Finding x and y #

Add the equations and read the total off.

Adding the two equations can hand over x + y without either letter being found

Two equations, and the question asks only for x + y, not for either letter.

Add them. Both letters land on 5, because 3 and 2 make 5 either way round.

Factor and divide: x + y = 4, with x and y each still unknown.

Solving in full gives x = 3 and y = 1 — the same 4, several steps later.

Inequalities #

Like an equation, until you divide by a negative.

An inequality is solved like an equation until you divide by a negative

Every value right of 4 works. The circle is hollow: 4 is not greater than 4.

Add the line under the sign and 4 joins in. Filled circle, same ray, one more value.

Solve 2x > 8 the way you solve an equation: halve both sides, x > 4. Same ray.

True: 2 < 3. Make both negative: −2 > −3. Bigger became smaller — the sign flips.

Two-Step Inequalities #

Two moves, one flip rule.

A two-step inequality solves like an equation, flipping only on a negative divide

Two steps stand between x and the answer: adding 3, then multiplying by 2.

Solve it like an equation: clear the 3, divide by 2. The sign holds its ground.

The answer is a ray: everything left of 4 works, and hollow 4 stays out.

One warning survives: dividing by a negative flips the sign. Only that step does.

Integer Solutions of Inequalities #

List who lives between the ends.

A compound inequality pens x between two ends, and the integers inside can be listed

Two ends at once: x sits right of −3, and no further up than 5. The overlap is it.

Hollow at −3: not allowed in. Filled at 5: the $\le$≤ lets 5 itself belong.

The integers inside run −2, −1, 0, 1, 2, 3, 4, 5 — the −3 is out, the 5 is in.

Count them off: eight integers. Check the ends first — they decide who belongs.

Forming an Inequality in Two Variables #

At most, at least, and the sign each one takes.

A limit on a total of two different things is written with an inequality sign

Pens at 2 and pads at 5 cost 2p + 5d, and 30 caps what that may reach.

At most and no more than give $\le$≤. At least and no fewer than give $\ge$≥.

5 pens and 4 pads cost exactly 30, and at most 30 lets the ceiling in.

A floor reads the same way: at least 20 items altogether is $p+d\ge 20$p + d ≥ 20.

Graphing an Inequality in Two Variables #

Draw the boundary, test a point, shade the side.

The solutions of an inequality in two letters fill one whole side of its boundary line

Start with the boundary. Every point on this line makes y = x + 1 exactly.

Test a point off the line. At (−2, 4): 4 beats −1. At (3, 0): 0 does not beat 4.

The side holding (−2, 4) is the solution, and the dashed boundary is left out.

Add the line under the sign and the boundary joins in, drawn solid.

Shading is not always upward. Test (0, 0): $0\le 6$0 ≤ 6 holds, so the origin side wins.

The Overlap of Two Inequality Regions #

Both must hold, so only the overlap counts.

A pair of inequalities is solved by the region where both shadings overlap

One inequality shades this half: every point in it satisfies $y\ge x-1$y ≥ x − 1.

The other shades this half, everything at or under the line y = 5 − x.

Draw both. The doubly shaded wedge is where the two demands hold at once.

Only that wedge solves the pair. (1, 2) is inside: $2\ge 0$2 ≥ 0 and $2\le 4$2 ≤ 4 both hold.

(4, 0) obeys $y\le 5-x$y ≤ 5 − x and breaks $y\ge x-1$y ≥ x − 1, so one out of two leaves it out.

Solving Absolute Value Equations #

The bars hide which way the inside came out.

A modulus equation splits into two cases, because two numbers sit at every distance

Two numbers sit five steps from zero, so |x| = 5 has two answers: 5 and −5.

The bars sit round x − 3, so the inside can be 5 or −5 — solve both.

Read that on the line: 8 and −2 are each five steps away from 3.

Both answers check out, because the bars report the same size either way.

A distance is never below zero, so |x + 1| = −4 has no answer at all.

Solving Absolute Value Inequalities #

One stretch, or two rays heading apart.

A modulus inequality either traps x between two values or splits it into two rays

Ask which numbers sit less than four steps from zero: that is |x| < 4.

The bars measure the distance from 3, so keep x within 5 steps of 3.

One stretch, sitting on 3 and reaching 5 each way: from −2 up to 8.

In general |x − a| < b means a − b < x < a + b, a stretch of width 2b.

Turn the sign round and you want the numbers further than 5 steps from 3.

Those are two rays heading apart, and no single stretch can describe them.

Solution Sets of Inequalities #

A region on the line, written as a set.

An inequality answers with a set — a region on the line, written in set-builder notation

2x + 1 > 7 solves to x > 3 — not one answer but a set: every number past 3.

Set-builder notation writes that set down: {x : x > 3}, read "all x such that".

{x : $x\ge -1$x ≥ −1} keeps its endpoint — the filled dot says −1 itself belongs.

Both bounds in one set: {x : $-1<x\le 4$−1 < x ≤ 4}, open at −1, closed at 4.