QuanticaAlgebra IOpen in the course4.3 ended with a promise, equations that hide the familiar both-sides shape behind fractions and heavier parentheses. This lesson keeps it. Nothing below is new mathematics. Every equation here becomes a 4.2 or 4.3 equation after one cleanup move you already own, and the lesson is learning to see which move each shape calls for.
Problem
Nothing in 4.3 had parentheses wrapped around both sides. Solve \(5(2x-3)=3(3x+4)\) anyway. Distribute on each side the 2.1 way, then collect the 4.3 way, and check in the original before entering the solution.
Show a hint
- The parentheses are a 2.1 job, not a new rule. Nothing crosses the equals sign until both sides are clean.
- Distributing on both sides gives \(10x-15=9x+12\).
Show the full solution
Distribute on each side, \(10x-15=9x+12\). Subtract \(9x\), \(x-15=12\), then add \(15\), \(x=\boxed{27}\). Check, \(5(54-3)=255\) and \(3(81+4)=255\). The parentheses only made the equation look unfamiliar, every move was an old one.
Problem
Solve \(7x-2(3x-8)=25\). The subtracted \(2\) distributes over both terms of \(3x-8\), signs included. Enter the solution.
Show a hint
- The minus in front of the parentheses belongs to everything inside, both terms.
- \(-2(3x-8)\) is \(-6x+16\), not \(-6x-16\).
Show the full solution
Distribute, \(7x-6x+16=25\), so \(x+16=25\) and \(x=\boxed{9}\). Check, \(63-2(27-8)=63-38=25\). Dropping the sign flip on the \(16\) is the single most common slip in this shape.
Fractions are the main event. \(\frac{x}{3}+\frac{x}{5}=16\) could be worked the 2.5 way, combining the left side into one fraction, but that drags fractions through every line. 4.2 allows multiplying both sides by any nonzero number, and one well-chosen multiplication removes every fraction at once. Pick a number every denominator divides.
Problem
Solve \(\frac{x}{3}+\frac{x}{5}=16\) by multiplying both sides by \(15\), then check in the original before entering the solution.
Show a hint
- At the solution both sides are the same number, and multiplying both copies by \(15\) keeps them equal.
- Multiplying both sides by \(15\) gives \(5x+3x=240\).
Show the full solution
Multiply both sides by \(15\), \(5x+3x=240\), so \(8x=240\) and \(x=\boxed{30}\). Check, \(\frac{30}{3}+\frac{30}{5}=10+6=16\). Multiplying by \(15\) cleared both fractions in one step.
Problem
Solve \(\frac{x}{6}+\frac{x}{9}=10\). The product \(54\) clears both denominators, and so does the LCD \(18\), with smaller numbers. Use \(18\) and enter the solution.
Show a hint
- \(18\) works because both \(6\) and \(9\) divide it.
- Multiplying both sides by \(18\) gives \(3x+2x=180\).
Show the full solution
Multiply both sides by \(18\), \(3x+2x=180\), so \(5x=180\) and \(x=\boxed{36}\). Check, \(6+4=10\). The \(54\) route gives \(9x+6x=540\) and the same answer, with numbers three times the size.
Problem
Solve \(\frac{x}{6}+5=\frac{x}{4}\). The multiplier \(12\) hits every term, the lone \(5\) included. Enter the solution.
Show a hint
- The \(5\) is a full term of the left side, and the multiplication distributes over the whole side.
- Clearing with \(12\) gives \(2x+60=3x\).
Show the full solution
Multiply both sides by \(12\), \(2x+60=3x\), so subtracting \(2x\) leaves \(x=\boxed{60}\). Check, \(10+5=15\) and \(\frac{60}{4}=15\). Leaving the \(5\) unmultiplied is the error this lesson's checks keep catching.
Problem
Solve \(\frac{x+9}{2}-\frac{x}{5}=6\). The fraction bar groups its numerator, so when you multiply by \(10\) the whole \(x+9\) travels together, in parentheses. Enter the solution.
Show a hint
- \(10\cdot\frac{x+9}{2}\) is \(5(x+9)\), never \(5x+9\).
- Clearing with \(10\) gives \(5(x+9)-2x=60\).
Show the full solution
Multiply both sides by \(10\), \(5(x+9)-2x=60\), so \(3x+45=60\), \(3x=15\), and \(x=\boxed{5}\). Check, \(\frac{14}{2}-1=7-1=6\). The bar is a grouping symbol, doing the same job parentheses do.
When each side of an equation is a single fraction and nothing else, two multiplications clear everything, one for each denominator, and each denominator cancels on its own side. Watch what remains after each cancellation, since the same pattern appears every time this shape does.
Problem
Solve \(\frac{2x+3}{9}=\frac{x+7}{6}\). Multiply both sides by a common multiple of the denominators so each one cancels. Enter the solution.
Show a hint
- The least common multiple of \(9\) and \(6\) is \(18\), so multiply both sides by \(18\).
- Clearing the denominators leaves \(2(2x+3)=3(x+7)\).
Show the full solution
Multiply both sides by \(18\), \(2(2x+3)=3(x+7)\), so \(4x+6=3x+21\) and \(x=\boxed{15}\). Check, \(\frac{33}{9}\) and \(\frac{22}{6}\) both equal \(\frac{11}{3}\).
Problem
Use the shortcut on \(\frac{2x+5}{12}=\frac{x-3}{9}\). The shortcut applies here, since each side is a single fraction and nothing else. Enter the solution as a fraction in lowest terms.
Show a hint
- Each numerator multiplies the other denominator, with whole numerators kept in parentheses.
- The cleared equation is \(9(2x+5)=12(x-3)\).
Show the full solution
Cross-multiply, \(9(2x+5)=12(x-3)\), so \(18x+45=12x-36\), \(6x=-81\), and \(x=\boxed{-\frac{27}{2}}\). Check, both sides come out \(-\frac{11}{6}\). A negative fraction solution is a normal outcome, exactly as in 4.3.
A decimal coefficient is a count of tenths or hundredths, so a decimal equation is a fraction equation. Tenths clear when both sides are multiplied by \(10\), hundredths need \(100\), and the finest decimal place present picks the power. The move is 4.2's same multiplication, and it hits every term, whole-number constants included.
Problem
Solve \(0.8x-3.1=1.7\) by multiplying both sides by \(10\) first, so every count of tenths becomes a whole number. Enter the solution.
Show a hint
- Every term is a count of tenths, so multiplying by \(10\) makes every coefficient whole.
- The cleared equation is \(8x-31=17\).
Show the full solution
Multiply both sides by \(10\), \(8x-31=17\), so \(8x=48\) and \(x=\boxed{6}\). Check, \(4.8-3.1=1.7\).
Problem
Solve \(0.06x+4=0.11x+1.5\). Hundredths call for \(100\), and the multiplication turns the whole \(4\) into \(400\). Enter the solution.
Show a hint
- Choose the multiplier by the finest decimal place present, and remember it reaches the plain \(4\) too.
- Clearing with \(100\) gives \(6x+400=11x+150\).
Show the full solution
Multiply both sides by \(100\), \(6x+400=11x+150\), so \(250=5x\) and \(x=\boxed{50}\). Check, \(3+4=7\) and \(5.5+1.5=7\). Multiplying by \(10\) instead would leave \(1.1x\) behind, which is why the finest place picks the power.
Problem
A student solves \(\frac{x}{2}-7=\frac{x}{6}+4\) by multiplying only the two fractions by \(6\), writes \(3x-7=x+4\), and gets \(x=\frac{11}{2}\). That candidate fails the 4.1 check in the original. Multiply every term by \(6\) and enter the correct solution.
Show a hint
- A legal multiplication reaches all four terms, the \(7\) and the \(4\) included.
- The correct cleared equation is \(3x-42=x+24\).
Show the full solution
Multiply every term by \(6\), \(3x-42=x+24\), so \(2x=66\) and \(x=\boxed{33}\). Check, \(16.5-7=9.5\) and \(5.5+4=9.5\). The student's check against the flawed line \(3x-7=x+4\) would have passed, which is why a check belongs to the original.
Every multiplier so far was a known nonzero number. In \(\frac{44}{x+15}=4\) the natural multiplier is \(x+15\), which contains the variable. 4.2's rule still holds wherever \(x+15\) is not zero, ruling out only \(x=-15\), where the original divides by zero anyway. Clear it, solve, and check in the original, which also shows the denominator was not zero.
Problem
Solve \(\frac{44}{x+15}=4\). Multiply both sides by \(x+15\), finish with 4.2's moves, and check your value in the original before entering it.
Show a hint
- After the one multiplication, the equation is a shape 4.2 solved many times.
- Clearing gives \(44=4(x+15)\), so \(44=4x+60\).
Show the full solution
Multiply both sides by \(x+15\), \(44=4(x+15)\), so \(44=4x+60\), \(4x=-16\), and \(x=\boxed{-4}\). Check, \(-4+15=11\) and \(\frac{44}{11}=4\), so the value works and the denominator was not zero. The check did logical work here, not just hygiene.
Every equation in this lesson stopped looking unfamiliar after one move. Distribute, clear the denominators, or clear the decimals, and each landed in 4.2's or 4.3's hands. 4.5 starts from equations with no algebra written down at all, plain English sentences, and turning those into equations is the next skill.
Practice these ideas
Practice
Solve \(4(2x+1)=3(x+8)\).
Show the solution
Distribute on both sides, \(8x+4=3x+24\). Subtract \(3x\), then subtract \(4\), so \(5x=20\) and \(x=\boxed{4}\). Check, both sides equal \(36\).
Practice
Solve \(9x-4(2x-7)=31\).
Show the solution
Distribute, \(9x-8x+28=31\), so \(x+28=31\) and \(x=\boxed{3}\). Check, \(27-4(-1)=31\). Dropping the sign and writing \(-28\) is the slip to watch here.
Practice
Solve \(\frac{4x+3}{5}=7\).
Show the solution
Multiply both sides by \(5\), \(4x+3=35\), so \(4x=32\) and \(x=\boxed{8}\). Check, \(\frac{35}{5}=7\).
Practice
Solve \(\frac{x}{3}+\frac{x}{7}=20\).
Show the solution
Multiply both sides by \(21\), \(7x+3x=420\), so \(10x=420\) and \(x=\boxed{42}\). Check, \(14+6=20\).
Practice
Solve \(\frac{x}{2}-\frac{x}{6}=13\).
Show the solution
Multiply both sides by \(6\), \(3x-x=78\), so \(2x=78\) and \(x=\boxed{39}\). Check, \(\frac{39}{2}-\frac{13}{2}=13\).
Practice
Solve \(0.9x-2.5=3.8\) by multiplying both sides by \(10\) first.
Show the solution
Multiply both sides by \(10\), \(9x-25=38\), so \(9x=63\) and \(x=\boxed{7}\). Check, \(6.3-2.5=3.8\).
Practice
Solve \(\frac{x}{6}=\frac{15}{9}\).
Show the solution
Cross-multiply, \(9x=6\cdot 15=90\), so \(x=\boxed{10}\). Check, both sides equal \(\frac{5}{3}\).
Practice
Solve \(\frac{x}{9}+4=\frac{x}{6}\). The multiplier reaches the \(4\) too.
Show the solution
Multiply both sides by \(18\), \(2x+72=3x\), so \(x=\boxed{72}\). Check, \(8+4=12\) and \(\frac{72}{6}=12\). Multiplying only the fractions is the error the check would have caught.
Practice
Solve \(\frac{x+8}{7}=\frac{x-4}{3}\).
Show the solution
Cross-multiply, \(3(x+8)=7(x-4)\), so \(3x+24=7x-28\), then \(4x=52\) and \(x=\boxed{13}\). Check, both sides equal \(3\).
Practice
Solve \(0.13x=0.05x+2\).
Show the solution
Multiply both sides by \(100\), \(13x=5x+200\), so \(8x=200\) and \(x=\boxed{25}\). Check, \(3.25=1.25+2\).
Practice
Solve \(\frac{x}{3}+8=\frac{x}{2}+5\).
Show the solution
Multiply both sides by \(6\), \(2x+48=3x+30\). Subtract \(2x\), then subtract \(30\), so \(x=\boxed{18}\). Check, \(6+8=14\) and \(9+5=14\).
Practice
Solve \(\frac{x}{8}+\frac{3}{4}=\frac{x}{2}\). Three fractions, so cross-multiplying does not apply. One multiplier clears all three.
Show the solution
Multiply both sides by \(8\), \(x+6=4x\), so \(3x=6\) and \(x=\boxed{2}\). Check, \(\frac{1}{4}+\frac{3}{4}=1\).
Practice
Solve \(\frac{x}{2}+\frac{x}{3}=-15\).
Show the solution
Multiply both sides by \(6\), \(3x+2x=-90\), so \(5x=-90\) and \(x=\boxed{-18}\). Check, \(-9-6=-15\).
Practice
Solve \(\frac{2x+7}{5}=\frac{x+1}{2}+3\). The extra \(+3\) means the cross-multiplying shortcut does not apply.
Show the solution
Multiply both sides by \(10\), \(2(2x+7)=5(x+1)+30\), so \(4x+14=5x+35\) and \(x=\boxed{-21}\). Check, \(\frac{-35}{5}=-7\) and \(\frac{-20}{2}+3=-7\).
Practice
A student solves \(0.5x-2=0.25x+4.5\) by multiplying only the \(x\) terms by \(100\), writes \(50x-2=25x+4.5\), and gets a candidate that fails the check in the original. Multiply every term by \(100\) and enter the correct solution.
Show the solution
Multiply every term by \(100\), \(50x-200=25x+450\), so \(25x=650\) and \(x=\boxed{26}\). Check, \(13-2=11\) and \(6.5+4.5=11\). The student's line skipped the \(2\) and the \(4.5\), so the two sides were scaled unequally and the candidate fails the original.
Practice
How many solutions does \(\frac{6x+9}{3}=2x+5\) have? Enter the count.
Show the solution
The left side is \(\frac{6x+9}{3}=2x+3\), so the equation reads \(2x+3=2x+5\). Subtracting \(2x\) leaves \(3=5\), which is false, so there are \(\boxed{0}\) solutions. The fraction was hiding a 4.3 outcome, not changing it.
Practice
Solve \(\frac{91}{x-6}=7\), and check that your value is not the excluded \(6\).
Show the solution
Multiply both sides by \(x-6\), \(91=7(x-6)\), so \(x-6=13\) and \(x=\boxed{19}\). Check, \(\frac{91}{13}=7\), and \(19\) is not the excluded \(6\), so the denominator was never zero.
Practice
Solve \(\frac{3x-2}{4}-\frac{x-3}{6}=2\). Enter the solution as a fraction in lowest terms.
Show the solution
Multiply both sides by \(12\), \(3(3x-2)-2(x-3)=24\), so \(9x-6-2x+6=24\), then \(7x=24\) and \(x=\boxed{\frac{24}{7}}\). Check, \(\frac{29}{14}-\frac{1}{14}=\frac{28}{14}=2\). Distributing the \(-2\) over both terms of \(x-3\) is where this one is usually lost.