QuanticaPrealgebraOpen in the courseAn expression like \(3x + 5\) just describes a whole family of values, one for each \(x\). Put an equals sign between two expressions, like \(3x + 5 = 20\), and you make a claim that both sides come out to the same number. That claim is an equation, and the value of the variable that makes it true is a solution. Two pictures help through this lesson. The equals sign is the level beam of a two-pan balance, and the variable is wrapped in layers of operations that you take off one at a time.
Problem
A vending machine charges \(3\) tokens to wake up plus \(5\) tokens per snack, so \(w\) snacks cost \(5w + 3\). One kid spent exactly \(23\) tokens, giving \(5w + 3 = 23\). With no method yet, test whole numbers. At \(w = 3\) the left side is \(18\), at \(w = 5\) it is \(28\), so the answer sits between. The value that makes both sides equal is the solution. What whole number \(w\) makes \(5w + 3 = 23\)? Type that value.
Show a hint
- You already passed \(23\) between \(w=3\) (which gives \(18\)) and \(w=5\) (which gives \(28\)). The answer is the whole number sitting right between them, so test the one you skipped.
- Try \(w=4\). Then \(5w\) is \(5\times 4=20\), and \(20+3=23\). That hits the target exactly, so that is your solution.
Show the full solution
Try \(w=4\). Then \(5\times 4=20\), and \(20+3=23\), which is exactly the right side, so \(w=\boxed{4}\). Testing values works here because the left side climbs by \(5\) each step, so once \(23\) is bracketed between \(w=3\) and \(w=5\) only one whole number is left to check.
Problem
Before solving an equation you should be able to check one. Substituting means replacing every \(x\) with a candidate and seeing whether both sides match. A note claims \(x = 7\) solves \(3x - 5 = 19\), but \(3(7) - 5 = 16 \neq 19\), so \(7\) fails. One of the values \(7\) or \(8\) actually solves \(3x - 5 = 19\). Type the value that works.
Show a hint
- A correct solution must make the left side, \(3x - 5\), come out to exactly \(19\). You already saw \(7\) gives \(16\), which is too small, so test the other candidate.
- Substitute \(x = 8\). The left side becomes \(3(8) - 5\). Work out \(3 \times 8 = 24\), then \(24 - 5\), and see whether you land on \(19\).
Show the full solution
Substitute \(x=8\). The left side is \(3(8)-5=24-5=19\), which matches the right side, so the value that works is \(\boxed{8}\). The other candidate fails, since \(3(7)-5=16\). Checking a value is always this direct, work out each side on its own and see whether they land on the same number.
Problem
A thermos starts at \(t\) degrees, the afternoon adds \(17\), and the display settles at \(9\), so \(t + 17 = 9\). The variable \(t\) is sealed inside an add-\(17\) box. Lift those \(17\) off the left pan, and lift the same \(17\) off the right pan to keep the scale level. Solve \(t + 17 = 9\) and type the value of \(t\).
Show a hint
- The \(17\) is added to \(t\). To undo an addition and free the variable, do the opposite operation to both sides at once. What single move strips that \(+17\) off the left pan?
- Subtract \(17\) from both sides. The left pan becomes just \(t\), and the right pan becomes \(9 - 17\). That difference dips below zero, which is fine since the thermos started cold.
Show the full solution
Subtract \(17\) from both sides. $$t + 17 = 9$$ $$t + 17 - 17 = 9 - 17$$ $$t = \boxed{-8}$$ Checking, \(-8 + 17 = 9\), which matches the right side. A negative answer is fine here, it just means the thermos started below zero.
Problem
Six identical sandbags, each of unknown mass \(m\), together tip a balance to \(15\) units, so \(6m = 15\). The only thing wrapped around \(m\) is a times-\(6\), and division undoes multiplication, so divide both sides by \(6\) to leave one bag alone. The answer need not be a whole number, which is fine. Solve \(6m = 15\) and type \(m\) as a fraction in lowest terms.
Show a hint
- The variable is multiplied by \(6\), so to peel that away you do the opposite of multiplying. Divide both sides of the equation by \(6\).
- Dividing both sides by \(6\) gives \(m = \frac{15}{6}\). Now reduce that fraction. Both \(15\) and \(6\) share a factor of \(3\), so cancel it to reach lowest terms.
Show the full solution
Divide both sides by \(6\). $$\frac{6m}{6} = \frac{15}{6}$$ The left side leaves \(m\) alone. On the right, \(15\) and \(6\) are both divisible by \(3\), so \(\frac{15}{6}\) reduces and each sandbag has mass $$\boxed{5/2}$$ Checking, \(6 \cdot \frac{5}{2} = \frac{30}{2} = 15\). A solution does not have to be a whole number.
Problem
In \(4x - 9 = 7\), the variable sits inside two boxes, since it was multiplied by \(4\) first, then had \(9\) taken away. To unwrap, peel in reverse and pull the outer box off first, so add \(9\) to both sides, then divide by \(4\). Solve \(4x - 9 = 7\) and type the value of \(x\).
Show a hint
- The outer layer is the minus \(9\), so deal with it first. What can you add to both sides to undo subtracting \(9\)?
- Add \(9\) to both sides to get \(4x = 16\). Now the variable sits in only the times-\(4\) box, so divide both sides by \(4\).
Show the full solution
Add \(9\) to both sides. $$4x - 9 + 9 = 7 + 9$$ That gives \(4x = 16\), so divide both sides by \(4\). $$\frac{4x}{4} = \frac{16}{4}$$ $$x = \boxed{4}$$ The \(-9\) came off first because it was the last thing done to \(x\). Checking, \(4(4) - 9 = 16 - 9 = 7\).
Problem
Start with \(n\), multiply by \(-3\), then add \(2\), and you reach \(10\), so \(-3n + 2 = 10\). The times-\(-3\) is the inner box and the add-\(2\) is the outer box, so peel from the outside in. Subtract \(2\) first, then divide by \(-3\). Watch the signs, because the answer will not be a whole number. Solve \(-3n + 2 = 10\) and type \(n\) as a fraction in lowest terms, including its sign.
Show a hint
- The outer box is the \(+2\), since the \(2\) was added last. Strip it off first by subtracting \(2\) from both sides, and see what is left on each side.
- After subtracting, you have \(-3n = 8\). The inner box is times \(-3\), so undo it by dividing both sides by \(-3\). A positive divided by a negative is negative, and \(8\) and \(3\) share no common factor, so it stays as a fraction.
Show the full solution
Subtract \(2\) from both sides to get \(-3n = 8\), then divide both sides by \(-3\). $$\frac{-3n}{-3} = \frac{8}{-3}$$ A positive over a negative is negative, and \(8\) and \(3\) share no common factor, so \(n = \boxed{-8/3}\). Checking, \(-3 \cdot \left(-\tfrac{8}{3}\right) = 8\) and \(8 + 2 = 10\). A negative fraction is a perfectly good solution.
Real equations are not always neat. Sometimes the variable's side holds several terms, like \(2x + 3x\), or a group that still needs distributing, like \(4(t - 7)\). Tidy that side first, using the like-terms and distributing work from 6.1, before you peel off a single layer. In every equation here the variable stays on one side and the other side is a plain number.
Problem
A delivery rider's day reads \(5k + 2 + 3k - 7 = 19\), where \(k\) is the number of trips. Do not undo anything yet. First tidy the cluttered left side by combining like terms, exactly as in 6.1, then isolate \(k\). Solve \(5k + 2 + 3k - 7 = 19\) and type the value of \(k\).
Show a hint
- Resist undoing anything first. The left side is a pile of like terms, so combine them before you touch the right side. Add up the \(k\) terms, \(5k\) and \(3k\), and separately add up the plain numbers, \(2\) and \(-7\).
- After combining you get \(8k - 5 = 19\), a clean two-step. Add \(5\) to both sides to get \(8k = 24\), then divide both sides by \(8\).
Show the full solution
Combine like terms on the left. The \(k\) terms give \(5k + 3k = 8k\), and the plain numbers give \(2 - 7 = -5\). $$8k - 5 = 19$$ Add \(5\) to both sides for \(8k = 24\), then divide both sides by \(8\). The rider made \(\boxed{3}\) trips. Tidying first turns a four-term pile into a plain two-step equation, so there is less to keep track of.
Problem
A side can hide its real shape inside a group. In \(6(2 - q) = 30\), the outer times-\(6\) has to reach both pieces inside before anything else can happen, so it meets the \(2\) and the \(-q\), and that minus sign rides along. Distribute carefully, then peel the two-step equation. Solve \(6(2 - q) = 30\) and type the value of \(q\).
Show a hint
- Hand the outer \(6\) to both pieces inside the group before you do anything else. The \(6\) hits the \(2\) and it also hits the \(-q\), so watch that the sign rides along. The left side turns into \(12 - 6q\).
- Now you have \(12 - 6q = 30\), a plain two-step. Subtract \(12\) from both sides to get \(-6q = 18\), then divide both sides by \(-6\). A positive divided by a negative is negative.
Show the full solution
Distribute the \(6\) across the group. $$6(2 - q) = 30$$ $$12 - 6q = 30$$ Subtract \(12\) from both sides to get \(-6q = 18\), then divide both sides by \(-6\). $$q = \frac{18}{-6} = \boxed{-3}$$ Checking, \(2 - (-3) = 5\) and \(6 \times 5 = 30\). The minus sign rides along when you distribute, so \(-q\) becomes \(-6q\), not \(6q\).
Problem
A paint mixer uses two thirds of a tank to fill today's order, so \(\frac{2}{3}x = 10\), where \(x\) is the tank's capacity in liters. The variable is wrapped in a multiply-by-\(\frac{2}{3}\) box, so multiply both sides by the reciprocal \(\frac{3}{2}\), which turns the \(\frac{2}{3}\) into a plain \(1\). Solve \(\frac{2}{3}x = 10\) and type the value of \(x\).
Show a hint
- The variable is being multiplied by \(\frac{2}{3}\). To undo a multiplication you do the opposite. Multiplying both sides by the reciprocal \(\frac{3}{2}\) will flip the \(\frac{2}{3}\) back to \(1\) and leave \(x\) alone on the left.
- Multiply both sides by \(\frac{3}{2}\). The left side becomes \(x\). The right side becomes \(10 \cdot \frac{3}{2}\), so work out \(\frac{30}{2}\).
Show the full solution
Multiply both sides by the reciprocal \(\frac{3}{2}\). $$\frac{3}{2} \cdot \frac{2}{3}x = \frac{3}{2} \cdot 10$$ On the left, \(\frac{3}{2} \cdot \frac{2}{3} = 1\), leaving \(x\). On the right, \(10 \cdot \frac{3}{2} = \frac{30}{2} = 15\), so the tank holds \(\boxed{15}\) liters. Multiplying both sides by \(3\) and then dividing by \(2\) gets you the same place, since that is the same two moves.
Problem
Solving for the variable is often just the first half of a question. A ticket booth's group passes \(x\) satisfy \(5(x - 2) = 35\), but the manager wants the wristband count, given by \(2x + 1\). First solve \(5(x - 2) = 35\) for \(x\), then compute \(2x + 1\). Type that final number.
Show a hint
- Start by solving for \(x\) on its own. The whole left side is multiplied by \(5\), so undo that first by dividing both sides by \(5\). Set the value of \(x\) aside, because the question is really asking about \(2x+1\).
- Dividing both sides by \(5\) gives \(x-2=7\), so adding \(2\) to both sides gives \(x=9\). Now substitute into the wristband rule. Work out \(2\times 9+1\) and type that single number.
Show the full solution
Divide both sides by \(5\) to get \(x - 2 = 7\), then add \(2\) to both sides for \(x = 9\). The question wants the wristband count, so substitute into \(2x + 1\). $$2(9) + 1 = 18 + 1 = 19$$ The booth hands out \(\boxed{19}\) wristbands. Solving for the variable was only half the job, so reread what the question actually asks for before typing an answer.
Problem
Sometimes you are handed the variable and must hunt for a missing number. A game designer's win happens when a score \(x\) satisfies \(3x + c = 20\), where \(c\) is a fixed bonus she still has to choose. She wants \(x = 4\) to count as a win, which means \(x = 4\) must be a solution. Substitute \(4\) for \(x\), then solve for \(c\). Type the value of \(c\).
Show a hint
- You know what \(x\) is this time, so put it to work. A solution is a value that makes the equation true, so replace \(x\) with \(4\) in \(3x + c = 20\) and see what the equation turns into.
- After substituting you get \(12 + c = 20\). Now \(c\) is the only unknown left, sitting on one side. Undo the \(+12\) by subtracting \(12\) from both sides.
Show the full solution
Substitute \(4\) in for \(x\), since \(x = 4\) has to make the equation true. $$3(4) + c = 20$$ $$12 + c = 20$$ Subtract \(12\) from both sides. $$c = \boxed{8}$$ Checking, \(3(4) + 8 = 12 + 8 = 20\). Nothing new is going on here. The letter you solve for is just whichever one is still unknown.
Problem
A community garden shares a total across \(4\) equal plots, ending with \(6\) beds each, so \(\dfrac{3(m-2)+2m}{4} = 6\), where \(m\) is beds per row. Peel the outermost box first by multiplying both sides by \(4\), then distribute, combine like terms, and isolate \(m\). Solve \(\dfrac{3(m-2)+2m}{4} = 6\) and type the value of \(m\).
Show a hint
- The whole left side is wrapped in a division by 4. That is the outermost box, so undo it first. Multiply both sides by 4 to clear the denominator, and the equation becomes \(3(m-2)+2m=24\).
- Now distribute the 3 across \(m-2\) to get \(3m-6\), then add the \(2m\). The like terms \(3m\) and \(2m\) combine into \(5m\), so the left collapses to \(5m-6=24\). Add 6 to both sides, then divide both sides by 5.
Show the full solution
Multiply both sides by \(4\) to clear the denominator. $$3(m-2)+2m=24$$ Distribute the \(3\), then combine the like terms \(3m\) and \(2m\). $$3m-6+2m=24$$ $$5m-6=24$$ Add \(6\) to both sides for \(5m=30\), then divide both sides by \(5\). $$m=\boxed{6}$$ The denominator was the outermost layer, so it had to come off before there was anything to tidy inside. Checking, \(3(4)+12=24\) and \(24 \div 4 = 6\).
Practice these ideas
Practice
A diver's instrument reads higher as she rises, and her start reading \(x\) satisfies \(x + 12 = 5\), where negative values mean below the surface. Solve \(x + 12 = 5\) and type the value of \(x\).
Show the solution
Subtract \(12\) from both sides. $$x + 12 = 5$$ $$x + 12 - 12 = 5 - 12$$ $$x = \boxed{-7}$$ Checking, \(-7 + 12 = 5\), which matches the right side. Landing below zero is fine, it just means she started below the surface.
Practice
A jar holds some marbles, and after \(6\) more are dropped in it holds \(20\). With \(n\) the starting count, that is \(n + 6 = 20\). Solve for \(n\) and type the value.
Show the solution
Subtract \(6\) from both sides. $$n + 6 - 6 = 20 - 6$$ $$n = 14$$ The jar started with \(\boxed{14}\) marbles. Checking, \(14 + 6 = 20\).
Practice
Four identical packs together hold \(9\) mini sets, so \(4m = 9\), where \(m\) is how many fit in one pack. Solve for \(m\) and type the value as a fraction in lowest terms.
Show the solution
Divide both sides by 4, which gives \(m = \tfrac{9}{4}\). Since 9 and 4 share no common factor, that is already in lowest terms, so \(\boxed{9/4}\). Checking, \(4 \cdot \tfrac{9}{4} = 9\).
Practice
A descending drone's tracker satisfies \(-6y = 54\). Divide both sides by \(-6\) to solve. Type the value of \(y\), including its sign.
Show the solution
Divide both sides by \(-6\). That leaves \(y\) on the left, and on the right \(54\) divided by \(6\) is \(9\), while a positive divided by a negative is negative. So \(y = \boxed{-9}\). Checking, \(-6 \times (-9) = 54\), which matches the right side.
Practice
A vending machine charges \(x\) per snack plus a flat \(\$7\) fee, and \(3\) snacks come to \(\$22\), so \(3x + 7 = 22\). Peel the constant, then the coefficient. Solve and type the value of \(x\).
Show the solution
Subtract \(7\) from both sides. $$3x + 7 - 7 = 22 - 7$$ $$3x = 15$$ Then divide both sides by \(3\). $$x = \boxed{5}$$ Checking, \(3(5) + 7 = 15 + 7 = 22\). The constant comes off before the coefficient, because adding the \(7\) was the last thing done.
Practice
A baker uses \(w\) cups on each of two trays and sets aside \(3\), using \(4\) cups in all, so \(2w - 3 = 4\). Add \(3\) to both sides, then divide by \(2\). Type \(w\) as a fraction in lowest terms.
Show the solution
Add 3 to both sides to get \(2w = 7\), then divide both sides by 2. Each tray takes \(\boxed{7/2}\) cups. Checking, \(2 \cdot \frac{7}{2} = 7\) and \(7 - 3 = 4\), which matches the right side.
Practice
Tidy the busy left side first, then hunt for the value. Solve \(7t - 2 - 4t + 5 = 18\) and type the value of \(t\).
Show the solution
Combine like terms on the left. The \(t\) terms give \(7t - 4t = 3t\), and the numbers give \(-2 + 5 = 3\). $$3t + 3 = 18$$ Subtract \(3\) from both sides for \(3t = 15\), then divide both sides by \(3\). $$t = \boxed{5}$$ Checking, \(7 \cdot 5 - 2 - 4 \cdot 5 + 5 = 35 - 2 - 20 + 5 = 18\).
Practice
Five identical kits each ship with \(4\) items missing, so each holds \(d - 4\), and all five total \(15\), giving \(5(d - 4) = 15\). Distribute the \(5\) first, then isolate \(d\). Type the value of \(d\).
Show the solution
Distribute the 5 across the parentheses. $$5(d-4)=15$$ $$5d-20=15$$ Add 20 to both sides for \(5d = 35\), then divide both sides by 5. $$d=\boxed{7}$$ Checking, each kit holds \(7 - 4 = 3\) items, and five kits hold \(5 \cdot 3 = 15\).
Practice
A triangle has three equal edges, each measuring \(2k + 5\), and its perimeter is \(9\), so \(3(2k + 5) = 9\). Distribute, then solve. Type the value of \(k\).
Show the solution
Distribute the \(3\) across the group. $$3(2k+5)=9$$ $$6k+15=9$$ Subtract \(15\) from both sides for \(6k = -6\), then divide both sides by \(6\). $$k=\boxed{-1}$$ Checking, each edge is \(2(-1)+5=3\), and three edges of \(3\) give a perimeter of \(9\).
Practice
Three fifths of a tank's capacity \(x\) comes to \(9\) liters, so \(\frac{3}{5}x = 9\). Multiply both sides by the reciprocal \(\frac{5}{3}\) to free \(x\). Type the value of \(x\).
Show the solution
Multiply both sides by the reciprocal \(\frac{5}{3}\). $$\frac{5}{3} \cdot \frac{3}{5}x = \frac{5}{3} \cdot 9$$ The two fractions cancel to \(1\) on the left, and on the right \(\frac{5}{3} \cdot 9 = \frac{45}{3} = 15\), so \(x = \boxed{15}\). Checking, \(\frac{3}{5} \cdot 15 = \frac{45}{5} = 9\).
Practice
A fractional coefficient scales \(b\), and the other side is negative, so \(\tfrac{2}{7}b = -4\). Free \(b\) by multiplying both sides by the flipped fraction \(\tfrac{7}{2}\), carrying the negative through. Type the value of \(b\).
Show the solution
Multiply both sides by the reciprocal \(\tfrac{7}{2}\). $$\tfrac{7}{2}\cdot\tfrac{2}{7}b = \tfrac{7}{2}\cdot(-4)$$ The left side collapses to \(b\), and the right side is \(\tfrac{-28}{2}\), so \(b = \boxed{-14}\). Checking, \(\tfrac{2}{7}\cdot(-14) = \tfrac{-28}{7} = -4\). The minus sign carries straight through the multiplication.
Practice
A value \(3h + 2\) shared equally among \(4\) display cases gives \(5\) each, so \(\dfrac{3h+2}{4} = 5\). Undo the division first by multiplying both sides by \(4\), then solve. Type the value of \(h\).
Show the solution
Multiply both sides by \(4\) to clear the denominator. $$3h+2=20$$ Subtract \(2\) from both sides for \(3h = 18\), then divide both sides by \(3\). $$h=\boxed{6}$$ Checking, \(3 \cdot 6 + 2 = 20\) and \(\frac{20}{4} = 5\). The denominator is the outermost layer, so it comes off before anything on the top.
Practice
A quantity \(5x + 2\) divided into \(4\) equal parts gives \(6\) each, so \(\frac{5x+2}{4} = 6\). Clear the denominator first by multiplying both sides by \(4\), then solve. Type \(x\) as a fraction in lowest terms.
Show the solution
Multiply both sides by \(4\) to clear the denominator. $$5x+2 = 24$$ Subtract \(2\) from both sides for \(5x = 22\), then divide both sides by \(5\). $$x = \boxed{22/5}$$ Since \(22\) and \(5\) share no common factor, that is already in lowest terms. Checking, \(5 \cdot \frac{22}{5} + 2 = 24\) and \(\frac{24}{4} = 6\).
Practice
You are told \(n = 5\) is a solution of \(2n + c = 17\), so plugging in \(5\) must make it true. Substitute \(n = 5\), then solve for the constant \(c\). Type the value of \(c\).
Show the solution
Substitute 5 in for \(n\), since \(n = 5\) has to make the equation true. $$2(5) + c = 17$$ $$10 + c = 17$$ Subtract 10 from both sides. $$c = \boxed{7}$$ Checking, \(2(5) + 7 = 10 + 7 = 17\), which matches the right side.