Equations & Inequalities

Study Sheet

Equations & Inequalities

Solving one-step and two-step equations and inequalities

What an Equation Is

Concept
The Big Idea

An equation is a math sentence that says two things are equal. It always has an equals sign, ==, with an expression on each side, like x+5=12x + 5 = 12. A letter such as xx is a variable, an unknown number we want to find. To solve an equation means to find the value of the variable that makes the sentence true. That value is called the solution.

Solving an equation is peeling operations off the variable in reverse order, doing the same thing to both sides each time. Two-step equations undo the addition first and the multiplication second — the reverse of the order in which they were applied. Inequalities are solved the same way, with one twist: multiplying or dividing by a negative flips the sign.

Substitution is the final judge of any solution, and it is worth seeing why. An equation makes a claim; a solution is a value that makes the claim true. So no matter how an answer was found, by algebra, by guessing, or by a table, putting it back into the ORIGINAL equation is the only test that decides whether it is right.

Checking Whether a Value Is a Solution

You can check a possible solution by substituting it in for the variable and seeing if both sides come out equal. If they match, it is a solution. If they do not match, it is not.

Example
Is x=4x = 4 a solution of x+5=12x + 5 = 12?

Substitute 44 for xx and simplify the left side:

x+5=124+5=12?9≠12\begin{aligned} x + 5 &= 12\\ 4 + 5 &= 12 \quad ?\\ 9 &\ne 12 \end{aligned}

The two sides are not equal, so x=4x = 4 is not a solution. (The real solution is x=7x = 7, since 7+5=127 + 5 = 12.)

Side note
Where it goes wrong

Checking an answer in a later line of your own work instead of the original equation. If a step was wrong, the later line is wrong too, and the "check" will agree with the mistake. Always substitute into the equation exactly as it was given.

Tip

Tip: Checking is never wasted work. After you solve any equation, substitute your answer back in. If both sides are equal, you know you are right.

Try it
Try it: which one is the solution?

Which of x=2x = 2, x=3x = 3, x=4x = 4 is a solution of 3x−1=113x - 1 = 11?

Show the answer

Substitute each value. x=2x = 2: 3(2)−1=5≠113(2) - 1 = 5 \ne 11. x=3x = 3: 3(3)−1=8≠113(3) - 1 = 8 \ne 11. x=4x = 4: 3(4)−1=113(4) - 1 = 11, which matches. So x=4x = 4 is the solution.

Inverse Operations and Keeping Balance

Concept
Undo with the Opposite

Every operation has an inverse (an opposite) that undoes it:

  • Addition and subtraction undo each other.
  • Multiplication and division undo each other.

To get the variable alone, use the inverse of whatever is happening to it.

Why does doing the same thing to both sides never lose the solution? If x+5x + 5 and 1212 are the same number, then subtracting 55 from each still leaves two equal numbers. The equation changes its appearance but not the set of values that make it true. Inverse operations are simply the choice of "same thing" that strips an operation off the variable.

Keep the Equation Balanced

Think of an equation like a balanced scale. The == sign is the center. Whatever you do to one side, you must do to the other side, or the scale tips and the equation is no longer true. Do the same operation to both sides every time.

Example
Balancing in Action
xx + 5 = 12: take 5 off each pan, leaving x = 7
A level scale is a true equation; removing 55 from both pans keeps it level.

To undo “+5+5” in x+5=12x + 5 = 12, subtract 55 from both sides:

x+5=12x+5−5=12−5x=7\begin{aligned} x + 5 &= 12\\ x + 5 - 5 &= 12 - 5\\ x &= 7 \end{aligned}

Subtracting 55 on both sides keeps the scale balanced and leaves xx alone.

Side note
Where it goes wrong

Subtracting 55 from only the left side: x+5−5=12x + 5 - 5 = 12, so "x=12x = 12". The scale tips. Check: 12+5=17≠1212 + 5 = 17 \ne 12. Whatever is removed from one pan must be removed from the other.

Tip

Golden rule: Do the same thing to both sides. Always. This one idea is the heart of solving every equation and inequality.

Try it
Try it: pick the inverse

Name the inverse operation that isolates the variable, then solve: (a) y−9=4y - 9 = 4; (b) 6w=426w = 42.

Show the answer

(a) 99 is subtracted from yy, so add 99 to both sides: y=13y = 13. Check: 13−9=413 - 9 = 4. (b) ww is multiplied by 66, so divide both sides by 66: w=7w = 7. Check: 6⋅7=426 \cdot 7 = 42.

One-Step Equations

Concept
One Move to Solve

A one-step equation needs just a single inverse operation to get the variable alone. Look at what is being done to the variable, then do the opposite to both sides.

Each one-step equation is solved by a single inverse because addition and subtraction undo each other exactly (x−8+8=xx - 8 + 8 = x for every xx), and so do multiplying and dividing by the same nonzero number (4x4=x\tfrac{4x}{4} = x). The zero exception matters: dividing by 00 is not allowed, which is why an equation like 0⋅x=50 \cdot x = 5 cannot be solved this way.

Example
Addition and Subtraction Equations

Solve x−8=3x - 8 = 3 (the 88 is subtracted, so add 88):

x−8=3x−8+8=3+8x=11\begin{aligned} x - 8 &= 3\\ x - 8 + 8 &= 3 + 8\\ x &= 11 \end{aligned}

Solve n+6=15n + 6 = 15 (the 66 is added, so subtract 66):

n+6=15n+6−6=15−6n=9\begin{aligned} n + 6 &= 15\\ n + 6 - 6 &= 15 - 6\\ n &= 9 \end{aligned}
Side note
Where it goes wrong

Solving y3=6\tfrac{y}{3} = 6 by dividing: y=2y = 2. The yy is already divided by 33, so the inverse is to multiply: y=18y = 18. Check: 183=6\tfrac{18}{3} = 6, while 23≠6\tfrac23 \ne 6.

Example
Multiplication and Division Equations

Solve 4x=204x = 20 (the xx is multiplied by 44, so divide by 44):

4x=204x4=204x=5\begin{aligned} 4x &= 20\\ \frac{4x}{4} &= \frac{20}{4}\\ x &= 5 \end{aligned}

Solve y3=6\dfrac{y}{3} = 6 (the yy is divided by 33, so multiply by 33):

y3=6y3⋅3=6⋅3y=18\begin{aligned} \frac{y}{3} &= 6\\ \frac{y}{3}\cdot 3 &= 6 \cdot 3\\ y &= 18 \end{aligned}
Tip

Tip: 4x4x means “44 times xx.” To undo multiplication, divide. Never subtract the 44; that will not free the variable.

Try it
Try it: four quick solves

Solve (a) m+14=9m + 14 = 9; (b) −5k=35-5k = 35; (c) r4=−3\dfrac{r}{4} = -3.

Show the answer

(a) Subtract 1414 from both sides: m=9−14=−5m = 9 - 14 = -5. (b) Divide both sides by −5-5: k=35−5=−7k = \tfrac{35}{-5} = -7. (c) Multiply both sides by 44: r=−12r = -12. Each checks: −5+14=9-5 + 14 = 9, −5(−7)=35-5(-7) = 35, −124=−3\tfrac{-12}{4} = -3.

Two-Step Equations

Concept
Undo in Reverse Order

A two-step equation has two operations, like 2x+3=112x + 3 = 11. Undo them in the reverse of the order of operations: first undo addition or subtraction, then undo multiplication or division. Peel the variable free from the outside in.

Why undo addition before multiplication? In 2x+32x + 3, the recipe was "double xx, then add 33." Taking it apart runs the recipe backward: remove the last thing done first. The tape diagram shows it: two copies of xx and 33 units make 1111; take away the 33 units and the two xx's make 88, so one xx is 44.

Example
Solve 2x+3=112x + 3 = 11
xx11111x + x + 3 = 11, so x + x = 8 and x = 4
Remove the 33 first, then split what is left into two equal parts.

First subtract 33 from both sides, then divide both sides by 22:

2x+3=112x+3−3=11−32x=82x2=82x=4\begin{aligned} 2x + 3 &= 11\\ 2x + 3 - 3 &= 11 - 3\\ 2x &= 8\\ \frac{2x}{2} &= \frac{8}{2}\\ x &= 4 \end{aligned}

Check: 2(4)+3=8+3=112(4) + 3 = 8 + 3 = 11. ✓

Side note
Where it goes wrong

Dividing first but only partly: from 2x+3=112x + 3 = 11 writing x+3=5.5x + 3 = 5.5. Dividing by 22 must divide every term: x+1.5=5.5x + 1.5 = 5.5, which still gives x=4x = 4. Subtracting first avoids the fractions entirely.

Example
Solve x5−2=4\dfrac{x}{5} - 2 = 4

First add 22 to both sides, then multiply both sides by 55:

x5−2=4x5=6x5⋅5=6⋅5x=30\begin{aligned} \frac{x}{5} - 2 &= 4\\ \frac{x}{5} &= 6\\ \frac{x}{5}\cdot 5 &= 6 \cdot 5\\ x &= 30 \end{aligned}

Check: 305−2=6−2=4\dfrac{30}{5} - 2 = 6 - 2 = 4. ✓

Tip

Tip: Order of operations builds an expression (multiply, then add). To take it apart, run backwards: undo the adding first, undo the multiplying last.

Try it
Try it: two steps

Solve 5−3x=175 - 3x = 17 and check.

Show the answer

The 55 is added to −3x-3x, so subtract 55 from both sides: −3x=12-3x = 12. Then divide by −3-3: x=−4x = -4. Check: 5−3(−4)=5+12=175 - 3(-4) = 5 + 12 = 17.

Combining Like Terms First

Concept
Tidy Up, Then Solve

Sometimes a side has more than one term with the same variable. Like terms have the same variable, so you can combine them: 3x+2x=5x3x + 2x = 5x. Simplify each side first, then solve as a normal equation.

Combining first is allowed because it does not change either side's value; it only rewrites it. 3x+2x3x + 2x is the same quantity as 5x5x for every xx, so replacing one with the other leaves the equation, and its solutions, untouched. The payoff is that the equation is back in a shape you already know how to solve.

Example
Solve 3x+2x=203x + 2x = 20

Combine the like terms on the left, then divide:

3x+2x=205x=205x5=205x=4\begin{aligned} 3x + 2x &= 20\\ 5x &= 20\\ \frac{5x}{5} &= \frac{20}{5}\\ x &= 4 \end{aligned}
Side note
Where it goes wrong

Reading 4x−x4x - x as 44. The lone xx has coefficient 11, so 4x−x=3x4x - x = 3x. With x=3x = 3: 4(3)−3=94(3) - 3 = 9, which is 3x3x, not 44.

Example
Solve 4x+3−x=124x + 3 - x = 12

Combine 4x4x and −x-x into 3x3x, then solve the two-step equation:

4x+3−x=123x+3=123x=9x=3\begin{aligned} 4x + 3 - x &= 12\\ 3x + 3 &= 12\\ 3x &= 9\\ x &= 3 \end{aligned}
Tip

Tip: xx by itself means 1x1x, so 4x−x=3x4x - x = 3x (not 44). Only combine terms that share the same variable.

Try it
Try it: tidy, then solve

Solve 7n−2−3n+8=267n - 2 - 3n + 8 = 26.

Show the answer

Combine the nn-terms: 7n−3n=4n7n - 3n = 4n. Combine the constants: −2+8=6-2 + 8 = 6. Now 4n+6=264n + 6 = 26, so 4n=204n = 20 and n=5n = 5. Check: 35−2−15+8=2635 - 2 - 15 + 8 = 26.

Writing Equations from Word Problems

Concept
Turn Words into Math

To solve a word problem, let a variable stand for the unknown, translate the words into an equation, then solve. Watch for signal words: “more than” or “increased by” means add; “less than” means subtract; “times” or “product” means multiply; “split” or “per” often means divide.

An equation is a sentence about the situation written in symbols, so it has to say something true about the story: the two sides must measure the same thing in the same units (here, dollars). Writing "let mm = miles" first forces you to decide what the letter counts, and that decision is what makes every other term fall into place.

Example
A Two-Step Word Problem

A taxi charges a $33 flat fee plus $22 per mile. A ride cost $1717. How many miles was it? Let mm be the number of miles. The cost is the fee plus $22 times the miles:

2m+3=172m=14m=7\begin{aligned} 2m + 3 &= 17\\ 2m &= 14\\ m &= 7 \end{aligned}

The ride was 77 miles. Check: 2(7)+3=172(7) + 3 = 17. ✓

Side note
Where it goes wrong

Writing 3m+2=173m + 2 = 17 for the taxi: that charges $3\$3 per mile and a $2\$2 fee, the reverse of the story. Test the setup on an easy case: a 00-mile ride must cost the flat fee, $3\$3, and 2(0)+3=32(0) + 3 = 3 does, while 3(0)+2=23(0) + 2 = 2 does not.

Tip

Tip: Always say in words what your variable stands for (“let mm = miles”). Then answer the question in a full sentence with units.

Try it
Try it: set up and solve

A gym charges a $25\$25 sign-up fee plus $15\$15 per month. Jada has paid $130\$130 in total. How many months has she been a member?

Show the answer

Let mm = number of months. The cost is the fee plus 1515 per month: 15m+25=13015m + 25 = 130. Subtract 2525: 15m=10515m = 105. Divide by 1515: m=7m = 7. She has been a member for 77 months. Check: 15(7)+25=13015(7) + 25 = 130.

What an Inequality Is

Concept
More Than One Answer

An inequality compares two expressions that are not necessarily equal. Instead of one solution, it usually has many. The four symbols are:

  • << means “less than”
  • >> means “greater than”
  • ≤\le means “less than or equal to”
  • ≥\ge means “greater than or equal to”

For example, x>5x > 5 means every number bigger than 55 is a solution.

An equation like x=3x = 3 picks out one point on the number line; an inequality like x≤3x \le 3 picks out a whole stretch of it. That is why inequality answers are drawn rather than listed: there are infinitely many solutions, including fractions like 2.52.5 and −12-\tfrac12, and a picture is the only honest way to show them all.

Example
Reading an Inequality
−2−10123456
Filled dot: 33 itself is a solution. The arrow covers everything below it.

The inequality x≤3x \le 3 means “xx is less than or equal to 33.” Solutions include 3, 2, 1, 0, −1,3,\ 2,\ 1,\ 0,\ -1, and every number below 33. The number 33 itself is allowed because of the “or equal to” part.

Side note
Where it goes wrong

Listing only whole numbers as the solutions of x≤3x \le 3: "3,2,1,03, 2, 1, 0". Every number to the left of 33 works, including 2.92.9, −12-\tfrac12 and −100-100.

Tip

Tip: The inequality sign always “opens” toward the larger side. In 7>27 > 2, the wide opening faces the bigger number, 77.

Try it
Try it: in or out?

Which of −4-4, 00, 55, 5.55.5, 66 are solutions of x>5x > 5?

Show the answer

A solution must be strictly greater than 55. −4-4, 00 and 55 are not (55 is equal, not greater). 5.55.5 and 66 are solutions.

Solving Inequalities

Concept
Solve Like an Equation, with One Warning

You solve inequalities almost exactly like equations: use inverse operations and do the same thing to both sides. One-step and two-step inequalities work the same way. There is just one special rule, coming next.

Adding or subtracting the same number slides both sides the same distance along the number line, so whichever side was on the right stays on the right. Multiplying or dividing by a positive number stretches or shrinks both sides by the same factor, which also keeps their order. Those are exactly the moves used here, and so the symbol never changes.

Example
One-Step Inequality: Solve x+4>9x + 4 > 9

Subtract 44 from both sides:

x+4>9x>5\begin{aligned} x + 4 &> 9\\ x &> 5 \end{aligned}

Every number greater than 55 is a solution.

Side note
Where it goes wrong

Stopping at 3x≤123x \le 12 and graphing "≤12\le 12". The variable is not alone until the 33 is divided out: x≤4x \le 4. Test x=10x = 10: 3(10)−2=283(10) - 2 = 28, which is not ≤10\le 10.

Example
Two-Step Inequality: Solve 3x−2≤103x - 2 \le 10
−101234567
The solution x≤4x \le 4: closed dot, arrow to the left.

Add 22 to both sides, then divide both sides by 33 (a positive number, so the sign does not change):

3x−2≤103x≤12x≤4\begin{aligned} 3x - 2 &\le 10\\ 3x &\le 12\\ x &\le 4 \end{aligned}

The solution is x≤4x \le 4.

Tip

Tip: Adding or subtracting any number, or multiplying/dividing by a positive number, keeps the inequality sign pointing the same direction.

Try it
Try it: solve and test

Solve 4x+7<194x + 7 < 19, then test one value inside and one outside your answer.

Show the answer

Subtract 77: 4x<124x < 12. Divide by 44 (positive, so no flip): x<3x < 3. Test inside, x=0x = 0: 4(0)+7=7<194(0) + 7 = 7 < 19, true. Test outside, x=5x = 5: 27<1927 < 19, false. The answer behaves as it should.

Flipping the Inequality Sign

Concept
The One Big Rule
−4−3−2−1012342 < 3−2 > −3
Multiplying by −1-1 mirrors the line: 2<32 < 3 becomes −2>−3-2 > -3.

When you multiply or divide both sides of an inequality by a negative number, you must flip the inequality sign. A “<<” becomes “>>,” and a “≤\le” becomes “≥\ge” (and the other way around). This is the single most important rule that makes inequalities different from equations.

Multiplying by −1-1 reflects the number line through 00: every number jumps to its mirror image on the other side. A mirror reverses left and right, so the number that was farther right ends up farther left. 33 is right of 22, but −3-3 is left of −2-2. Any negative multiplier includes this reflection, which is why the symbol must turn around; a positive multiplier only stretches, and order survives.

Example
Dividing by a Negative: Solve −2x<8-2x < 8

Divide both sides by −2-2 and flip << to >>:

−2x<8−2x−2>8−2x>−4\begin{aligned} -2x &< 8\\ \frac{-2x}{-2} &> \frac{8}{-2}\\ x &> -4 \end{aligned}

Why? Test x=0x = 0: −2(0)=0<8-2(0) = 0 < 8 is true, and 0>−40 > -4 is true. The flip keeps the answer correct.

Side note
Where it goes wrong

Forgetting to flip when dividing by a negative: from −2x<6-2x < 6 writing x<−3x < -3. Test x=0x = 0, which satisfies the original (0<60 < 6): it is not less than −3-3, so that answer throws away real solutions. The correct answer is x>−3x > -3.

Example
Two-Step with a Flip: Solve −3x+1≥7-3x + 1 \ge 7

Subtract 11 first (no flip), then divide by −3-3 and flip ≥\ge to ≤\le:

−3x+1≥7−3x≥6−3x−3≤6−3x≤−2\begin{aligned} -3x + 1 &\ge 7\\ -3x &\ge 6\\ \frac{-3x}{-3} &\le \frac{6}{-3}\\ x &\le -2 \end{aligned}
Tip

Remember: Only multiplying or dividing by a negative flips the sign. Adding or subtracting a negative does not flip it.

Try it
Try it: flip or not?

Solve (a) −4x≥20-4x \ge 20; (b) x−6<−9x - 6 < -9; (c) x−2>3\dfrac{x}{-2} > 3.

Show the answer

(a) Divide by −4-4 and flip: x≤−5x \le -5. (b) Only 66 is added, no multiplying: x<−3x < -3, no flip. (c) Multiply by −2-2 and flip: x<−6x < -6. Check (c) with x=−8x = -8: −8−2=4>3\tfrac{-8}{-2} = 4 > 3.

Graphing Inequality Solutions

Concept
Show the Answer on a Number Line

Because an inequality has many solutions, we picture them on a number line. Put a circle at the boundary number and shade the direction of all the solutions:

  • Use an open circle (hollow) for << or >>, because the boundary number is not included.
  • Use a closed circle (filled in) for ≤\le or ≥\ge, because the boundary number is included.

Then shade to the right for greater-than, or to the left for less-than.

The circle answers one question: is the boundary number itself a solution? For x>2x > 2, testing x=2x = 2 gives 2>22 > 2, false, so the dot is left open; for x≤−1x \le -1, testing x=−1x = -1 gives −1≤−1-1 \le -1, true, so it is filled. The arrow then covers every number that makes the inequality true, which is always one whole side of the boundary.

Example
Graphing x>2x > 2
−3−2−1012345
Open circle at 22, arrow to the right.

Draw a number line. Place an open circle on 22 (since 22 is not a solution), and shade the arrow to the right to show every number greater than 22.

Side note
Where it goes wrong

Shading toward the arrow of the symbol when the variable is on the right: 4>x4 > x does not mean "shade right." It says xx is less than 44, so shade left. Rewrite it with xx first, x<4x < 4, before drawing.

Example
Graphing x≤−1x \le -1
−5−4−3−2−10123
Closed circle at −1-1, arrow to the left.

Place a closed circle on −1-1 (since −1-1 is a solution), and shade to the left for all numbers less than −1-1.

Tip

Tip: Open circle == “not included” (<< or >>). Closed circle == “included” (≤\le or ≥\ge). The “or equal to” line under the symbol is your reminder to fill the circle in.

Try it
Try it: describe the graph

Describe the graph of −3≥x-3 \ge x: the circle at the boundary (open or closed) and the direction of the arrow.

Show the answer

Rewrite with xx first: x≤−3x \le -3. The boundary −3-3 is included, so the circle is closed, and the arrow points left.

Going Deeper: Advanced Equations & Inequalities

Variables on Both Sides and Clearing Fractions

Concept
Get the Variable onto One Side

When the variable appears on both sides, like 5x−3=2x+95x - 3 = 2x + 9, first gather all the variable terms onto one side and all the plain numbers onto the other. Add or subtract a variable term from both sides to move it, then finish with the usual two-step method. If the equation has fractions, you can clear the denominators first: multiply every term on both sides by the least common denominator (LCD). This turns a messy fraction equation into a clean whole-number one.

Moving a variable term across works for the same reason as moving a number: 2x2x is a quantity, and subtracting the same quantity from both sides keeps them equal. Clearing denominators multiplies both sides by one nonzero number, which is also a balance-preserving move; choosing the LCD simply makes every fraction cancel at once.

Example
Variables on Both Sides: Solve 5x−3=2x+95x - 3 = 2x + 9

Subtract 2x2x from both sides to collect the variable on the left, then finish:

5x−3=2x+95x−3−2x=2x+9−2x3x−3=93x=12x=4\begin{aligned} 5x - 3 &= 2x + 9\\ 5x - 3 - 2x &= 2x + 9 - 2x\\ 3x - 3 &= 9\\ 3x &= 12\\ x &= 4 \end{aligned}

Check: left =5(4)−3=17= 5(4) - 3 = 17; right =2(4)+9=17= 2(4) + 9 = 17. Both sides equal 1717. ✓

Example
Clearing Denominators: Solve x2+1=x3+3\dfrac{x}{2} + 1 = \dfrac{x}{3} + 3

The denominators are 22 and 33, so the LCD is 66. Multiply every term by 66:

x2+1=x3+36⋅x2+6⋅1=6⋅x3+6⋅33x+6=2x+18x+6=18x=12\begin{aligned} \frac{x}{2} + 1 &= \frac{x}{3} + 3\\ 6\cdot\frac{x}{2} + 6\cdot 1 &= 6\cdot\frac{x}{3} + 6\cdot 3\\ 3x + 6 &= 2x + 18\\ x + 6 &= 18\\ x &= 12 \end{aligned}

Check: left =122+1=7= \dfrac{12}{2} + 1 = 7; right =123+3=7= \dfrac{12}{3} + 3 = 7. ✓

Two Special Cases: No Solution and Infinitely Many

Concept
When the Variable Disappears

Sometimes, while solving, the variable cancels out completely and you are left with a statement of pure numbers. What that statement says tells you the answer:

  • If you get a false statement like 5=−35 = -3, there is no solution. No number could ever make the original equation true.
  • If you get a true statement like 8=88 = 8, there are infinitely many solutions. Every number works, because the two sides were really the same expression in disguise (this is called an identity).

For example, 2x+5=2x−32x + 5 = 2x - 3 gives 5=−35 = -3 after subtracting 2x2x: no solution. But 4x+8=4(x+2)4x + 8 = 4(x + 2) becomes 4x+8=4x+84x + 8 = 4x + 8, then 8=88 = 8: infinitely many solutions.

Tip

Tip: A normal equation lands on “x=x = (a number).” If instead the variable vanishes, stop and read what remains: false means no solution, true means all numbers.

A First Look at Systems of Two Equations

Concept
Two Equations, Two Unknowns

A system is two equations that must both be true at the same time, usually with two variables like xx and yy. The solution is the pair of values that satisfies both. One reliable method is substitution: if one equation already tells you what a variable equals (for example y=x+1y = x + 1), plug that expression into the other equation. This leaves a single equation in one variable, which you already know how to solve. Then back-substitute to find the second variable.

Example
Substitution: Solve the System y=x+1y = x + 1 and 2x+y=72x + y = 7
xy−11234−11234567(2, 3)y = x + 12x + y = 7
The two lines cross at (2,3)(2, 3): the one point on both.

The first equation says yy is x+1x + 1. Substitute x+1x + 1 in place of yy in the second equation:

2x+y=72x+(x+1)=73x+1=73x=6x=2\begin{aligned} 2x + y &= 7\\ 2x + (x + 1) &= 7\\ 3x + 1 &= 7\\ 3x &= 6\\ x &= 2 \end{aligned}

Now find yy using y=x+1=2+1=3y = x + 1 = 2 + 1 = 3. The solution is the pair (x,y)=(2,3)(x, y) = (2, 3). Check: y=x+1⇒3=2+1y = x + 1 \Rightarrow 3 = 2 + 1 ✓ and 2x+y=2(2)+3=72x + y = 2(2) + 3 = 7 ✓

Compound Inequalities and Interval Notation

Concept
Two Conditions at Once

A compound inequality traps the variable between two bounds, like −1<2x+3≤9-1 < 2x + 3 \le 9. It really means two inequalities joined by “and”: the middle is greater than −1-1 and at most 99. To solve, do the same operation to all three parts at once. Interval notation is a compact way to write the answer: a round bracket ( )(\ ) excludes an endpoint (matches << or >>), and a square bracket [ ][\ ] includes it (matches ≤\le or ≥\ge). So −2<x≤3-2 < x \le 3 is written (−2, 3](-2,\ 3].

Example
Solve −1<2x+3≤9-1 < 2x + 3 \le 9 and Write the Interval
−4−3−2−1012345
Open at −2-2, closed at 33: the interval (−2,3](-2, 3].

Subtract 33 from all three parts, then divide all three parts by 22 (positive, so no flip):

−1< 2x+3 ≤9−1−3< 2x ≤9−3−4< 2x ≤6−2< x ≤3\begin{aligned} -1 < \ &2x + 3 \ \le 9\\ -1 - 3 < \ &2x \ \le 9 - 3\\ -4 < \ &2x \ \le 6\\ -2 < \ &x \ \le 3 \end{aligned}

The solution is every number greater than −2-2 and up to and including 33. In interval notation: (−2, 3](-2,\ 3]. Check: x=3x = 3 gives 2(3)+3=9≤92(3) + 3 = 9 \le 9 ✓; x=−2x = -2 gives 2(−2)+3=−12(-2)+3 = -1, which is not >−1> -1, so −2-2 is correctly excluded. ✓

Tip

Remember: If you ever multiply or divide a compound inequality by a negative number, flip both signs at once. And match your brackets to your symbols: round for strict (<,><,>), square for “or equal” (≤,≥\le,\ge).

Why It Works

Adding the same number to both sides of a true equation keeps it true, and so does multiplying by a nonzero number; that is why the balance method is sound. For inequalities, multiplying by −1-1 reverses order because it reflects the number line: 2<32 < 3 but −2>−3-2 > -3. Every other step (adding, multiplying by a positive) preserves order, so only the negative-multiplier step needs the flip.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

Formulas, Proofs & Tips

Tip
Solving equations and inequalities
Do the same thing to both sides;multiplying an inequality by a negative flips it\text{Do the same thing to both sides};\qquad \text{multiplying an inequality by a negative flips it}

What it means. Equations stay balanced under identical operations; inequalities have one extra rule.

Example. 2x+3=11⇒2x=8⇒x=42x+3=11\Rightarrow 2x=8\Rightarrow x=4.

Why it works. If a=ba=b then a+c=b+ca+c=b+c and ac=bcac=bc — equality is preserved. But if a<ba<b, multiplying by −1-1 reverses their order on the number line, since −a-a now lies to the right of −b-b.

Tip. Undo operations in reverse PEMDAS order: addition first, multiplication last.