QuanticaAlgebra IOpen in the courseIn 4.1 a check settled any candidate. Substitute, evaluate both sides, compare. What it never did was produce a candidate. This lesson builds the producer, a way to trade an equation for a simpler equation with exactly the same solutions, and the first trade is one you can already make.
Problem
The equation \(w+24=63\) is true at exactly one value of \(w\), and at that value its two sides are the same number. Take \(24\) away from that number on each side. On the left this leaves \(w\) alone. Enter the number the right side becomes.
Show a hint
- At the solution both sides are one number. Taking the same amount from each copy of that number leaves equal amounts.
- Subtracting \(24\) from the left side leaves \(w\) alone. Do the identical subtraction to the \(63\).
Show the full solution
\(63-24=39\), so the right side becomes \(\boxed{39}\). The new equation reads \(w=39\), and the check \(39+24=63\) confirms it. That one subtraction solved the equation.
Problem
Solve \(t-38=26\). One balance move does it, adding \(38\) to both sides removes the subtraction and leaves \(t\) alone. Check your value in the original equation, the 4.1 way, before entering it.
Show a hint
- Undo the subtraction with the opposite move, made on both sides.
- Adding \(38\) to both sides leaves \(t\) alone. The right side becomes \(26+38\).
Show the full solution
Add \(38\) to both sides, \(t=26+38=\boxed{64}\). Check, \(64-38=26\). The check costs seconds and catches every slip.
The same argument covers scaling. At a solution both sides are one number, tripling that number on each side keeps the sides equal, and dividing by \(3\) undoes it. It looks like any multiplier works. One of them does not, and the next problem finds it.
Problem
The equation \(6d=84\) is true only at \(d=14\). Multiply both sides by \(0\) and every term collapses, leaving \(0=0\). Test the values \(-3\), \(0\), \(8\), \(14\), and \(20\) in the new equation. Enter how many of them are solutions of the new equation but not of the original.
Show a hint
- The new equation never mentions \(d\), so ask which values could possibly fail it.
- All five values satisfy \(0=0\), and the original is satisfied only at \(14\). Count the values that pass one equation but not the other.
Show the full solution
Every value makes \(0=0\) true, and only \(14\) satisfies \(6d=84\), so \(\boxed{4}\) of the five are new. Multiplying by zero manufactured solutions, and nothing undoes it, since dividing by zero is not a move.
Problem
Solve \(12m=132\). The left side is twelve copies of \(m\), so divide both sides by \(12\). Check the value before entering it.
Show a hint
- Division is the reverse of the multiplication that built the left side.
- Dividing both sides by \(12\) leaves \(m\) alone against \(132\) divided by \(12\).
Show the full solution
Divide both sides by \(12\), \(m=\frac{132}{12}=\boxed{11}\). Check, \(12\cdot 11=132\).
Problem
Solve \(\frac{n}{15}=8\) by multiplying both sides by \(15\). Check the value in the original equation before entering it.
Show a hint
- The left side divided \(n\) by \(15\), so undo it with the opposite scaling move.
- Multiplying both sides by \(15\) leaves \(n\) alone. The right side becomes \(8\) times \(15\).
Show the full solution
Multiply both sides by \(15\), \(n=8\cdot 15=\boxed{120}\). Check, \(\frac{120}{15}=8\).
In 4.1 you read \(9x+7=52\) with no method at all. The term \(9x\) had to be \(45\), and dividing by \(9\) finishes it, \(x=5\). That reading was really two balance moves, subtract \(7\) from both sides, then divide both sides by \(9\). The left side was built by multiplying and then adding, so it unwinds in the opposite order.
Problem
Solve \(3a+29=98\). Subtract first to expose the term \(3a\), then divide. Check your value in the original equation before entering it.
Show a hint
- The \(29\) sits outside the product \(3a\), so it is the last thing that happened and the first thing to undo.
- Subtracting \(29\) from both sides gives \(3a=69\).
Show the full solution
Subtract \(29\) from both sides, \(3a=69\), then divide by \(3\), \(a=\boxed{23}\). Check, \(3\cdot 23+29=98\).
Problem
Solve \(10y+74=82\). The moves are the same as before, and the answer is not a whole number. Enter it as a fraction in lowest terms.
Show a hint
- Nothing about the moves needs the answer to be whole. Unwind as usual.
- Subtracting \(74\) gives \(10y=8\). Divide and reduce.
Show the full solution
Subtract \(74\), \(10y=8\), divide by \(10\), \(y=\frac{8}{10}=\boxed{\frac{4}{5}}\). Check, \(10\cdot\frac{4}{5}+74=8+74=82\). Fractional solutions are normal outcomes, not signs of a mistake.
Sometimes a side needs cleanup before any unwinding. When the variable side is a pile of like terms, combine them the 2.4 way first. After that the equation is two-step and the moves are the ones you already have.
Problem
Solve \(16q+27-8q+13=100\). Combine like terms on the left before making any balance move. Enter the solution as a fraction in lowest terms.
Show a hint
- The left side is not two-step yet. Collect the \(q\) terms and the constants first.
- \(16q-8q=8q\) and \(27+13=40\), so the equation is \(8q+40=100\).
Show the full solution
Combining gives \(8q+40=100\), so \(8q=60\) and \(q=\frac{60}{8}=\boxed{\frac{15}{2}}\). Check, \(8\cdot\frac{15}{2}+40=60+40=100\).
Problem
Solve \(\frac{3}{4}c=51\). One move does it, multiply both sides by the reciprocal \(\frac{4}{3}\), the 2.5 skill. Enter the solution.
Show a hint
- A fractional coefficient clears in one move, scaling by the number that multiplies with it to \(1\).
- \(c=51\cdot\frac{4}{3}\). Divide \(51\) by \(3\) before multiplying.
Show the full solution
Multiply both sides by \(\frac{4}{3}\), \(c=51\cdot\frac{4}{3}=\boxed{68}\). Check, \(\frac{3}{4}\cdot 68=51\).
Problem
Solve \(28-x=93\). Subtracting \(28\) from both sides leaves \(-x=65\), a statement about the opposite of \(x\). One more move finishes it. Enter the solution.
Show a hint
- \(-x\) means \((-1)x\), and \(-1\) is a nonzero number like any other.
- Divide both sides of \(-x=65\) by \(-1\), or read it directly, the opposite of \(x\) is \(65\).
Show the full solution
Subtract \(28\), \(-x=65\), then multiply both sides by \(-1\), \(x=\boxed{-65}\). Check, \(28-(-65)=28+65=93\). If the opposite of \(x\) is \(65\), then \(x\) is \(-65\), the moves and the reading agree.
Problem
A student solves \(5x+30=80\) by erasing the \(30\) from the left side only, getting \(5x=80\) and then \(x=16\). Check that candidate the 4.1 way. Evaluate the left side at \(x=16\) and enter how much it overshoots the right side.
Show a hint
- A check needs no opinion about the moves. Substitute the candidate and evaluate the original left side.
- At \(x=16\) the left side is \(5(16)+30\). Compare that with \(80\).
Show the full solution
The left side at \(16\) is \(5\cdot 16+30=110\), and \(110-80=\boxed{30}\), exactly the \(30\) that left one side but not the other. Legal moves hit both sides. Subtracting \(30\) from both gives \(5x=50\), so \(x=10\), and \(5\cdot 10+30=80\) checks.
Problem
Solve \(\frac{3}{5}h+46=22\). Undo the addition first, then multiply by the reciprocal. The solution is negative, and the check still takes ten seconds. Enter the solution.
Show a hint
- Last operation first. The \(46\) was added after the coefficient did its work.
- Subtracting \(46\) gives \(\frac{3}{5}h=-24\). Multiply by \(\frac{5}{3}\).
Show the full solution
Subtract \(46\), \(\frac{3}{5}h=-24\), multiply by \(\frac{5}{3}\), \(h=-24\cdot\frac{5}{3}=\boxed{-40}\). Check, \(\frac{3}{5}(-40)+46=-24+46=22\).
Every equation in this lesson kept its variable on one side, and four reversible moves handled all of them. 4.3 starts where that stops, equations with the variable on both sides, and the balance principle carries over unchanged.
Practice these ideas
Practice
Enter the solution of \(x+34=59\).
Show the solution
Subtract \(34\) from both sides, \(x=59-34=\boxed{25}\). Check, \(25+34=59\), so the value passes the 4.1 test.
Practice
Enter the solution of \(y-27=44\).
Show the solution
Add \(27\) to both sides, \(y=44+27=\boxed{71}\). Check, \(71-27=44\).
Practice
Enter the solution of \(7k=119\).
Show the solution
Divide both sides by \(7\), \(k=\frac{119}{7}=\boxed{17}\). Check, \(7\cdot 17=119\).
Practice
Enter the solution of \(\frac{t}{9}=12\).
Show the solution
Multiply both sides by \(9\), \(t=12\cdot 9=\boxed{108}\). Check, \(\frac{108}{9}=12\).
Practice
Enter the solution of \(w+53=35\).
Show the solution
Subtract \(53\) from both sides, \(w=35-53=\boxed{-18}\). Check, \(-18+53=35\). Subtracting from both sides works the same even when the result is negative.
Practice
Enter the solution of \(2k+41=79\).
Show the solution
Subtract \(41\), \(2k=38\), then divide by \(2\), \(k=\boxed{19}\). Check, \(2\cdot 19+41=38+41=79\).
Practice
Enter the solution of \(6p+61=13\).
Show the solution
Subtract \(61\), \(6p=13-61=-48\), then divide by \(6\), \(p=\boxed{-8}\). Check, \(6(-8)+61=-48+61=13\).
Practice
Enter the solution of \(\frac{k}{12}+5=8\).
Show the solution
Subtract \(5\), \(\frac{k}{12}=3\), then multiply both sides by \(12\), \(k=\boxed{36}\). Check, \(\frac{36}{12}+5=3+5=8\).
Practice
Combine like terms first, then solve \(13w+12-8w+73=175\). Enter the solution.
Show the solution
Combining gives \(5w+85=175\), so \(5w=90\) and \(w=\boxed{18}\). Check, \(13\cdot 18+12-8\cdot 18+73=234+12-144+73=175\). Combining is not a balance move at all, it rewrites one side into an equal expression, so nothing on the other side changes.
Practice
Solve \(2(u+67)=232\), distributing once the 2.1 way or dividing both sides by \(2\) first. Enter the solution.
Show the solution
Divide both sides by \(2\), \(u+67=116\), then subtract \(67\), \(u=\boxed{49}\). Check, \(2(49+67)=2\cdot 116=232\). Distributing first gives \(2u+134=232\) and lands on the same value.
Practice
Enter the solution of \(\frac{2}{3}g=58\).
Show the solution
Multiply both sides by \(\frac{3}{2}\), \(g=58\cdot\frac{3}{2}=29\cdot 3=\boxed{87}\). Check, \(\frac{2}{3}\cdot 87=58\).
Practice
Enter the solution of \(-12u=72\).
Show the solution
Divide both sides by \(-12\), \(u=\frac{72}{-12}=\boxed{-6}\). Check, \(-12\cdot(-6)=72\). A positive product from a negative coefficient forces a negative solution.
Practice
Enter the solution of \(66-x=145\).
Show the solution
Subtract \(66\), \(-x=79\), then multiply both sides by \(-1\), \(x=\boxed{-79}\). Check, \(66-(-79)=66+79=145\).
Practice
Solve \(8n+77=86\). Enter the solution as a fraction in lowest terms.
Show the solution
Subtract \(77\), \(8n=9\), then divide by \(8\), \(n=\boxed{\frac{9}{8}}\). Check, \(8\cdot\frac{9}{8}+77=9+77=86\). A fractional answer is a normal outcome, not a sign of a slip.
Practice
To solve \(3x-21=99\), a student divides the \(3x\) and the \(99\) by \(3\) but leaves the \(-21\) alone, writing \(x-21=33\), so \(x=54\). Check that candidate the 4.1 way. Evaluate the left side of the original equation at \(x=54\) and enter how much it exceeds the right side.
Show the solution
The left side at \(54\) is \(3\cdot 54-21=162-21=141\), and \(141-99=\boxed{42}\). Dividing both sides means every term, \(x-7=33\), or add \(21\) first to get \(3x=120\). Either way \(x=40\), and \(3\cdot 40-21=99\) checks.
Practice
Combine like terms first, then solve \(3m+32-4m+57=96\). Enter the solution.
Show the solution
Combining gives \(-m+89=96\), so \(-m=7\) and \(m=\boxed{-7}\). Check, \(3(-7)+32-4(-7)+57=-21+32+28+57=96\). A net coefficient of \(-1\) reads as the opposite of \(m\), so \(m\) is the opposite of \(7\).
Practice
Solve \(\frac{8}{3}n=-14\). Enter the solution as a fraction in lowest terms.
Show the solution
Multiply both sides by \(\frac{3}{8}\), \(n=-14\cdot\frac{3}{8}=-\frac{42}{8}=\boxed{-\frac{21}{4}}\). Check, \(\frac{8}{3}\cdot\left(-\frac{21}{4}\right)=-\frac{168}{12}=-14\). Multiplying by the positive number \(\frac{3}{8}\) does not change the sign, so the solution stays negative.