QuanticaAlgebra IOpen in the courseThe choice between substitution and elimination comes down to coefficients, and a coefficient is only readable once an equation is written \(ax+by=c\). Plenty of systems are not written that way. Every system in this lesson is an ordinary one after a cleanup move from 4.4, and the new fact is that each of the two equations is cleaned on its own.
Problem
Eliminating \(y\) from \(6(x-2y)+7y=29\) and \(4x+9y=155\) means comparing the two coefficients of \(y\). In the second equation that coefficient is \(9\). Enter the coefficient of \(y\) in the first equation, once its left side is written as \(ax+by\).
Show a hint
- To eliminate \(y\) you compare the two coefficients of \(y\), and that coefficient is whatever multiplies \(y\) once the left side is collected into one \(x\)-term and one \(y\)-term.
- Distribute the \(6\) over \(x-2y\) to get \(6x-12y\), then combine that with the \(+7y\) outside the parentheses.
Show the full solution
Distributing gives \(6x-12y+7y\), and \(-12y+7y=-5y\), so the left side is \(6x-5y\) and the coefficient of \(y\) is \(\boxed{-5}\). The printed \(7\) is not the coefficient on its own, since distributing the \(6\) produces a second \(y\)-term, \(-12y\), that combines with it.
Problem
Distributing and combining like terms turns \(6(x-2y)+7y=29\) into \(6x-5y=29\), so the system is \(6x-5y=29\) and \(4x+9y=155\). Eliminate \(x\) the 5.3 way, back-substitute, and enter \(x\).
Show a hint
- Nothing about elimination changes now that both equations are in \(ax+by=c\) form. Compare the two coefficients on \(x\) before scaling anything.
- The \(x\)-coefficients \(6\) and \(4\) both divide \(12\), so double the first equation and triple the second.
Show the full solution
Doubling the first equation gives \(12x-10y=58\) and tripling the second gives \(12x+27y=465\). Subtracting the first of those from the second leaves \(37y=407\), so \(y=11\), and then \(6x-55=29\) gives \(6x=84\) and \(x=\boxed{14}\). Check, \(6(14)-5(11)=84-55=29\) and \(4(14)+9(11)=56+99=155\). Only the cleanup was new. Once both equations are in \(ax+by=c\) form, elimination is the same work as in 5.3.
Problem
Neither equation of \(4(x+3y)-5y=123\) and \(6x-3(y-4)=75\) is in \(ax+by=c\) form yet. Clean each one, solve the system, and enter the product \(xy\). In the second equation the factor in front of the parentheses is \(-3\), and it reaches both terms inside, signs included.
Show a hint
- Expand and combine on the left of each equation the 2.1 way before a solving method is picked. The constant that comes out of the second expansion belongs on the right side with the \(75\).
- \(-3(y-4)\) is \(-3y+12\), not \(-3y-12\), so the second equation cleans to \(6x-3y=63\) and then \(2x-y=21\). Isolating \(y\) there costs one move.
Show the full solution
Distributing gives \(4x+12y-5y=123\), so \(4x+7y=123\), and \(6x-3y+12=75\), so \(6x-3y=63\) and \(2x-y=21\). Then \(y=2x-21\), and \(4x+7(2x-21)=123\) gives \(18x=270\), \(x=15\), \(y=9\), and \(xy=\boxed{135}\). The \(-3\) reaches the \(-4\) as well, which is why the constant lands as \(+12\).
Problem
Neither \(9x-33=3y+132\) nor \(4y+5x=2x+20\) reads \(ax+by=c\), since one has a number on the left and each has letter terms on both sides. Use 4.3's balance moves on each, then solve the cleaned system and enter \(y\).
Show a hint
- Take one equation at a time. Move every letter term to the left side and every number to the right side, doing the same thing to both sides at each step.
- The first equation collects to \(9x-3y=165\), and dividing both sides by \(3\) gives \(3x-y=55\). The second collects to \(3x+4y=20\), so the \(x\)-columns match.
Show the full solution
Collecting the first equation gives \(9x-3y=165\), and dividing both sides by \(3\) gives \(3x-y=55\). The second collects to \(3x+4y=20\). Subtracting the cleaned first equation from the cleaned second leaves \(5y=-35\), so \(y=\boxed{-7}\), with \(3x=55-7=48\) and \(x=16\). Check \((16,-7)\) in both originals, \(144-33=111=-21+132\) and \(-28+80=52=32+20\). Once both equations read \(ax+by=c\), one subtraction finished the system.
Problem
Clean \(7x+4y=2x+3y+86\) and \(4(x+2y)-x-3y=144\) into \(ax+by=c\) form. Exactly one of the two letters can then be isolated with no division at all. Enter the value of that letter.
Show a hint
- Nothing can be compared until both equations are in \(ax+by=c\) form. Combine like terms in the first, and distribute the \(4\) in the second.
- The cleaned system is \(5x+y=86\) and \(3x+5y=144\). Scan its four coefficients for a \(1\).
Show the full solution
Combining like terms in the first equation gives \(5x+y=86\), and distributing in the second gives \(4x+8y-x-3y=144\), so \(3x+5y=144\). The only coefficient of \(1\) is on the \(y\) of the first equation, so \(y=86-5x\), and \(3x+5(86-5x)=144\) becomes \(430-22x=144\), so \(22x=286\) and \(x=13\). Then \(y=86-65=\boxed{21}\). Eliminating \(y\) means scaling one equation to \(5y\), and eliminating \(x\) means scaling both to \(15x\), so \(y\) is the cheap letter either way.
4.4 cleared a fractional equation by multiplying both sides by a common denominator of its fractions, and one well-chosen multiplication removed every fraction at once. A system has two equations to run that on. The next two problems put fractions in one equation, then in both.
Problem
Solve \(\frac{x}{3}+\frac{y}{2}=17\) and \(4x-3y=60\). Only one of the two equations has fractions in it, and the other already reads \(ax+by=c\). Enter \(x\).
Show a hint
- The second equation has no fractions in it, so it needs no multiplier at all. Leave it as it is and work on the first.
- The denominators \(3\) and \(2\) both divide \(6\). Multiply every term of the first equation by \(6\), the \(17\) included.
Show the full solution
Multiplying the first equation by \(6\) gives \(2x+3y=102\), and adding \(4x-3y=60\) leaves \(6x=162\), so \(x=\boxed{27}\), with \(3y=102-54=48\) and \(y=16\). Check, \(9+8=17\) and \(108-48=60\). Each equation is cleaned on its own, so only the fractional one is multiplied.
Problem
Solve \(\frac{x}{10}+\frac{y}{3}=15\) and \(\frac{x}{9}+\frac{y}{2}=19\). Each equation has its own pair of denominators, so clear the fractions one equation at a time. Enter \(x\).
Show a hint
- Clearing the first equation takes a multiple of \(10\) and \(3\). Clearing the second takes a multiple of \(9\) and \(2\). Choose the two multipliers separately.
- Multiply the first equation by \(30\) and the second by \(18\), every term on both sides.
Show the full solution
Multiplying the first equation by \(30\) gives \(3x+10y=450\), and multiplying the second by \(18\) gives \(2x+9y=342\). Doubling the first result gives \(6x+20y=900\), tripling the second gives \(6x+27y=1026\), and subtracting leaves \(7y=126\), so \(y=18\). Then \(3x+180=450\) gives \(3x=270\) and \(x=\boxed{90}\). The multiplier for each equation is built from that equation's own denominators, so the two multipliers are different.
Problem
A student clears \(\frac{x}{3}+4=\frac{y}{9}\) by multiplying by \(9\) and gets \(3x+4=y\). Paired with \(5x+y=124\) that line gives \((15,49)\), and the check fails, since \(\frac{15}{3}+4=9\) while \(\frac{49}{9}\) is not \(9\). Redo the clearing and enter the correct \(y\).
Show a hint
- Multiplying by \(9\) has to multiply every term, and the first equation has three terms, not two.
- The \(+4\) sits on the same side as \(\frac{x}{3}\), so it gets multiplied as well and becomes \(36\).
Show the full solution
Multiplying every term by \(9\) gives \(3x+36=y\). Substituting into \(5x+y=124\) gives \(5x+3x+36=124\), so \(8x=88\), \(x=11\), and \(y=3(11)+36=\boxed{69}\). Check, \(\frac{11}{3}+4=\frac{23}{3}\) and \(\frac{69}{9}=\frac{23}{3}\). Every term means the \(+4\) as well, so \(3x+4=y\) is a different equation, and the pair it gives fails the original.
A decimal coefficient is a count of tenths or hundredths, so a decimal equation is really a fraction equation, and one multiplication clears it just as before. The power of ten still comes from one equation at a time, so the two multipliers need not match.
Problem
Solve \(0.6x+0.4y=37\) and \(0.07x-0.05y=1.9\). The power of ten that clears an equation comes from the finest decimal place in that equation. Enter \(x\).
Show a hint
- The first equation has nothing finer than tenths, and the second has hundredths, so the two powers of ten are not the same.
- Multiply the first equation by \(10\) and the second by \(100\), both sides each time. The right sides then read \(370\) and \(190\).
Show the full solution
Clearing gives \(6x+4y=370\) and \(7x-5y=190\). Scaling to \(20y\) and \(-20y\) gives \(30x+20y=1850\) and \(28x-20y=760\), and adding leaves \(58x=2610\), so \(x=\boxed{45}\), with \(y=25\). Check, \(27+10=37\) and \(3.15-1.25=1.9\). Multiplying the second equation by \(10\) would leave \(0.7x\) and \(0.5y\) behind.
One shape from 4.4 is worth naming again. A fraction bar is a grouping symbol, so a numerator with two terms is grouped as a whole and needs parentheses the moment it is multiplied. Inside a system that numerator carries two letters instead of one, and that is the only difference from 4.4.
Problem
Solve \(\frac{4x-y}{5}=\frac{x+2y}{4}\) and \(x+y=125\). Each side of the first equation is a single fraction and nothing else, so two multiplications clear it. Enter \(x\).
Show a hint
- This is 4.4's shape with two letters in each numerator instead of one. The fraction bar groups its whole numerator.
- Multiplying by \(4\) and then by \(5\) leaves \(4(4x-y)=5(x+2y)\). Collect that into \(ax+by=c\) before using \(x+y=125\).
Show the full solution
Multiplying by \(4\) and then by \(5\) gives \(4(4x-y)=5(x+2y)\), so \(16x-4y=5x+10y\) and \(11x=14y\). With \(x=125-y\), \(11(125-y)=14y\) gives \(25y=1375\), \(y=55\), and \(x=\boxed{70}\). Check, \(70+55=125\), and both sides of the first equation come out \(45\), from \(\frac{225}{5}\) and \(\frac{180}{4}\). In \(ax+by=c\) form the cleaned equation is \(11x-14y=0\), so \(c\) is allowed to be \(0\).
Some systems are not linear in \(x\) and \(y\) at all. In \(\frac{9}{x}+\frac{2}{y}=41\) the letters are in the denominators, and multiplying through by \(xy\) produces an \(xy\) term rather than a linear equation. Both equations are linear in \(\frac{1}{x}\) and \(\frac{1}{y}\) though, and those two pieces can be given names.
Problem
In \(\frac{4}{x}+\frac{3}{y}=44\) and \(\frac{7}{x}-\frac{3}{y}=11\), set \(u=\frac{1}{x}\) and \(v=\frac{1}{y}\). Solve the system in \(u\) and \(v\) and enter \(u\).
Show a hint
- Rewrite every \(\frac{1}{x}\) as \(u\) and every \(\frac{1}{y}\) as \(v\), so \(\frac{4}{x}\) is \(4u\) and \(\frac{3}{y}\) is \(3v\).
- The renamed system is \(4u+3v=44\) and \(7u-3v=11\), an ordinary 5.3 problem with opposite \(v\)-coefficients.
Show the full solution
In \(u\) and \(v\) the system is \(4u+3v=44\) and \(7u-3v=11\). The \(v\)-coefficients are opposites, so adding gives \(11u=55\) and \(u=\boxed{5}\). Then \(20+3v=44\) gives \(v=8\), and since \(u=\frac{1}{x}\) and \(v=\frac{1}{y}\), the original values are \(x=\frac{1}{5}\) and \(y=\frac{1}{8}\).
Problem
Solve \(\frac{14}{x}+\frac{6}{y}=5\) and \(\frac{21}{x}-\frac{4}{y}=1\). Neither equation has the form \(ax+by=c\), but naming the two repeated pieces yourself turns both into that form. Solve the renamed system and enter \(x\) itself, not the value of the new letter.
Show a hint
- Neither equation is linear in \(x\) and \(y\), but both are linear in \(\frac{1}{x}\) and \(\frac{1}{y}\). Name those two pieces.
- With \(u=\frac{1}{x}\) and \(v=\frac{1}{y}\) the system is \(14u+6v=5\) and \(21u-4v=1\). Solve that by elimination, then convert with \(x=\frac{1}{u}\).
Show the full solution
With \(u=\frac{1}{x}\) and \(v=\frac{1}{y}\) the system is \(14u+6v=5\) and \(21u-4v=1\). Doubling the first and tripling the second gives \(28u+12v=10\) and \(63u-12v=3\), and adding leaves \(91u=13\), so \(u=\frac{1}{7}\) and \(v=\frac{1}{2}\). Then \(x=\frac{1}{u}=\boxed{7}\), with \(y=2\). Check in the originals, \(\frac{14}{7}+\frac{6}{2}=5\) and \(\frac{21}{7}-\frac{4}{2}=1\).
Every system in this lesson was one cleanup move away from an ordinary system, and the choice between substitution and elimination came after that move rather than before it. 5.5 starts further back. The equations are not written down at all, only English sentences, and building the system is the work.
Practice these ideas
Practice
The first equation of the system \(3(x+2y)=54\) and \(x-y=6\) is not in \(ax+by=c\) form yet. Rewrite it, solve the system, and enter \(x\).
Show the solution
Dividing the first equation by \(3\) gives \(x+2y=18\). Subtracting \(x-y=6\) leaves \(3y=12\), so \(y=4\). Back-substituting into \(x-y=6\) gives \(x=\boxed{10}\). Check, \(3(10+8)=54\) and \(10-4=6\). Distributing to \(3x+6y=54\) and then dividing by \(3\) lands on the same equation, so either route cleans it.
Practice
The system is \(8x-4(y-3)=52\) and \(2x+y=34\). Clean the first equation into \(ax+by=c\) form, then solve the system. The ask is \(y\), not \(x\).
Show the solution
Distributing gives \(8x-4y+12=52\), so \(8x-4y=40\) and \(2x-y=10\). Adding \(2x+y=34\) leaves \(4x=44\), so \(x=11\) and \(y=\boxed{12}\). Check \((11,12)\) in the original first equation, \(88-4(9)=52\). The product of \(-4\) and \(-3\) is \(+12\), and dropping that sign change is the usual slip here.
Practice
Both sides of \(9x-y=5x+3y+44\) carry letters. Put that equation into \(ax+by=c\) form, solve the system it makes with \(2x+5y=120\), and enter \(x\).
Show the solution
Collecting the letters on the left gives \(4x-4y=44\), and dividing by \(4\) leaves \(x-y=11\), so \(x=y+11\). Then \(2(y+11)+5y=120\) collects to \(7y=98\), so \(y=14\) and \(x=\boxed{25}\). Check, \(9(25)-14=211\) and \(5(25)+3(14)+44=211\). In \(x-y=11\) the letter \(x\) has coefficient \(1\), so isolating it brings in no fractions.
Practice
The equation \(12+3x=5y-8\) is not in \(ax+by=c\) form. Rewrite it in that form, solve it together with \(2x+3y=69\), and enter the value of \(y\).
Show the solution
Subtract \(12\) from both sides and \(5y\) from both sides to get \(3x-5y=-20\). Multiply that by \(2\) for \(6x-10y=-40\), multiply \(2x+3y=69\) by \(3\) for \(6x+9y=207\), and subtract the first of those from the second for \(19y=247\), so \(y=\boxed{13}\). Then \(3x=45\) and \(x=15\). In the original, \(12+45=57\) and \(65-8=57\).
Practice
Solve \(\frac{x}{4}+\frac{y}{5}=14\) and \(3x-2y=36\). Only one of the two equations has fractions to clear. Enter \(x\).
Show the solution
Multiplying the first equation by \(20\) gives \(5x+4y=280\). Doubling \(3x-2y=36\) gives \(6x-4y=72\), and adding leaves \(11x=352\), so \(x=\boxed{32}\). Back-substituting gives \(y=30\). Check, \(\frac{32}{4}+\frac{30}{5}=8+6=14\) and \(96-60=36\). Only the equation with denominators needs a multiplier, and it has to reach the right side, which is why the \(14\) becomes \(280\).
Practice
Solve \(\frac{x+2y}{3}-\frac{y}{4}=9\) and \(2x-y=26\). Multiply the first equation by the least common denominator to clear the fractions, then solve the cleaned system. Enter \(x\).
Show the solution
Multiplying the first equation by \(12\) gives \(4(x+2y)-3y=108\), so \(4x+5y=108\). The second equation gives \(y=2x-26\), so \(4x+10x-130=108\) and \(14x=238\), so \(x=\boxed{17}\). Then \(y=8\). Check in the first equation, \(\frac{17+16}{3}-\frac{8}{4}=11-2=9\). Multiply the whole numerator and not just the \(x\), which is why \(x+2y\) stays in parentheses until the distributing step.
Practice
Solve \(\frac{x}{2}+\frac{y}{7}=13\) and \(\frac{x}{8}+\frac{y}{5}=9\). The denominators differ from one equation to the other, so clear each equation with its own multiplier. Enter \(x\).
Show the solution
Multiplying the first equation by \(14\) gives \(7x+2y=182\), and the second by \(40\) gives \(5x+8y=360\). Four times the first is \(28x+8y=728\), and subtracting \(5x+8y=360\) leaves \(23x=368\), so \(x=\boxed{16}\). Then \(y=35\). Check, \(\frac{16}{8}+\frac{35}{5}=9\). The smallest multiplier for each equation comes from its own denominators.
Practice
Solve \(0.8x+0.5y=18\) and \(0.2x-0.3y=6.2\). Clear the decimals from each equation first, then eliminate. Enter \(y\).
Show the solution
Multiplying each equation by \(10\) gives \(8x+5y=180\) and \(2x-3y=62\). Four times the second is \(8x-12y=248\), and subtracting that from the first leaves \(17y=-68\), so \(y=\boxed{-4}\). Back-substitution gives \(x=25\). Check, \(0.8(25)+0.5(-4)=20-2=18\) and \(0.2(25)-0.3(-4)=5+1.2=6.2\).
Practice
Solve \(0.7x+0.4y=45\) and \(0.03x+0.08y=3.5\). Clear the decimals in each equation first, then solve. Enter \(x\).
Show the solution
Multiplying the first equation by \(10\) gives \(7x+4y=450\), and the second by \(100\) gives \(3x+8y=350\). Doubling \(7x+4y=450\) gives \(14x+8y=900\), and subtracting \(3x+8y=350\) leaves \(11x=550\), so \(x=\boxed{50}\). Then \(y=25\). Check in the second equation, \(0.03(50)+0.08(25)=3.5\). A multiplier of \(10\) there would leave \(0.3x\), since a hundredths place needs a multiplier of \(100\).
Practice
The system is \(0.25x+0.4y=17\) and \(x-y=3\). A student clears the first equation by multiplying by \(100\) and writes \(25x+40y=17\), leaving the right side alone. That line has no whole-number solution with \(x-y=3\) at all, which is the first sign something is wrong. Redo the clearing and enter the correct \(x\).
Show the solution
Multiplying every term by \(100\) gives \(25x+40y=1700\), and dividing by \(5\) gives \(5x+8y=340\). With \(x=y+3\), \(5(y+3)+8y=340\) gives \(13y=325\), so \(y=25\) and \(x=\boxed{28}\). Check, \(0.25(28)+0.4(25)=7+10=17\) and \(28-25=3\). Multiplying only the left side changes the equation, since \(25x+40y=17\) is the same as \(0.25x+0.4y=0.17\).
Practice
The system is \(\frac{x}{2}-6=\frac{y}{8}\) and \(2x+y=138\). A student multiplies the first equation by \(8\), writes \(4x-6=y\), and reports \(x=24\). Testing \((24,90)\) in the first equation gives \(6\) on the left and \(11.25\) on the right. Enter the correct value of \(x\).
Show the solution
Multiplying every term by \(8\) gives \(4x-48=y\), so \(4x-y=48\). Adding \(2x+y=138\) gives \(6x=186\), so \(x=\boxed{31}\). The student multiplied only the two fraction terms and left the \(-6\) unchanged. With \(y=76\), the check is \(15.5-6=9.5\) and \(\frac{76}{8}=9.5\).
Practice
Solve \(\frac{2x+y}{4}=\frac{x+5y}{6}\) and \(x+y=99\). Clean the first equation into \(ax+by=c\) before using the second. Enter \(y\).
Show the solution
Multiplying both sides by \(4\) and then by \(6\) gives \(6(2x+y)=4(x+5y)\), so \(12x+6y=4x+20y\) and the cleaned form is \(8x-14y=0\), or \(4x=7y\). With \(x=99-y\), \(4(99-y)=7y\) gives \(396-4y=7y\), so \(11y=396\) and \(y=\boxed{36}\). Then \(x=63\), and the original equation checks, \(\frac{162}{4}=\frac{243}{6}=40.5\). The fraction bar groups the whole numerator, so each numerator needs parentheses once the denominators are gone.
Practice
Solve \(\frac{x-y}{4}=\frac{y}{6}+3\) and \(x+y=52\). The right side of the first equation is a fraction plus a whole number rather than a single fraction, so clear it with one multiplier instead of cross-multiplying. Enter \(x\).
Show the solution
Multiplying the first equation by \(12\) gives \(3(x-y)=2y+36\), so \(3x-5y=36\). With \(x=52-y\), \(3(52-y)-5y=36\) gives \(156-8y=36\), so \(8y=120\), \(y=15\), and \(x=\boxed{37}\). Check, \(\frac{37-15}{4}=\frac{11}{2}\) and \(\frac{15}{6}+3=\frac{11}{2}\). One multiplication by \(12\) clears both denominators, and the \(3\) on the right is a term to multiply like any other.
Practice
Clean \(3(x+2y)-2y=71\) and \(8x+9y=2x+2y+131\) into \(ax+by=c\) form. One letter's two coefficients meet at a smaller least common multiple than the other letter's, so eliminating it takes less scaling. Enter that letter.
Show the solution
Distributing gives \(3x+6y-2y=71\), so \(3x+4y=71\), and collecting the second gives \(6x+7y=131\). The \(x\)-coefficients \(3\) and \(6\) meet at \(6\), which takes one scaling, while the \(y\)-coefficients \(4\) and \(7\) meet at \(28\), which takes two, so the cheaper letter is \(\boxed{x}\). Both routes reach the same pair \((9,11)\), and the only difference is how large the numbers get.
Practice
In \(\frac{1}{x}+\frac{5}{y}=33\) and \(\frac{3}{x}-\frac{1}{y}=19\), set \(u=\frac{1}{x}\) and \(v=\frac{1}{y}\). Solve the renamed system and enter \(u\).
Show the solution
Renaming gives \(u+5v=33\) and \(3u-v=19\). Solving the second for \(v\) gives \(v=3u-19\), so \(u+5(3u-19)=33\), which is \(16u-95=33\), then \(16u=128\) and \(u=\boxed{8}\). Then \(v=5\), and the check gives \(8+25=33\) and \(24-5=19\). Since \(u=\frac{1}{x}\) and \(v=\frac{1}{y}\), the original solution is \(x=\frac{1}{8}\) and \(y=\frac{1}{5}\).
Practice
Solve the system \(\frac{2}{x}+\frac{3}{y}=30\) and \(\frac{5}{x}-\frac{2}{y}=37\). Enter \(x\) as a fraction in lowest terms.
Show the solution
With \(u=\frac{1}{x}\) and \(v=\frac{1}{y}\) the system is \(2u+3v=30\) and \(5u-2v=37\). Doubling the first gives \(4u+6v=60\) and tripling the second gives \(15u-6v=111\), and the sum is \(19u=171\), so \(u=9\) and \(v=4\). Then \(y=\frac{1}{4}\) and \(x=\frac{1}{u}=\boxed{\frac{1}{9}}\). The ask is \(x\), not \(u\), so \(9\) is one step short of the answer.
Practice
Solve \(\frac{x+y}{2}+\frac{y}{3}=23\) and \(0.15x-0.2y=1.5\). Clear each equation to the form \(ax+by=c\) first. Enter the product \(xy\).
Show the solution
Multiplying the first equation by \(6\) gives \(3(x+y)+2y=138\), so \(3x+5y=138\). Multiplying the second by \(100\) gives \(15x-20y=150\), and dividing by \(5\) gives \(3x-4y=30\). Subtracting leaves \(9y=108\), so \(y=12\), and \(3x=138-60=78\), so \(x=26\) and \(xy=\boxed{312}\). Check, \(\frac{38}{2}+4=23\) and \(3.9-2.4=1.5\). Clearing fractions and clearing decimals are the same move, one multiplier taken from the denominators and one from a power of ten.
Practice
Nonnegative numbers \(a\) and \(b\) satisfy \(\sqrt{a}+3\sqrt{b}=41\) and \(\sqrt{a}-\sqrt{b}=13\). Neither equation is linear in \(a\) and \(b\), but both are linear in \(\sqrt{a}\) and \(\sqrt{b}\). Name those two pieces, solve, convert back, and enter \(a\).
Show the solution
With \(u=\sqrt{a}\) and \(v=\sqrt{b}\) the system is \(u+3v=41\) and \(u-v=13\). Subtracting gives \(4v=28\), so \(v=7\) and \(u=20\). Then \(a=u^2=\boxed{400}\). Squaring is what converts back, so \(b=49\), and the check gives \(20+21=41\) and \(20-7=13\). Entering \(20\) reports \(\sqrt{a}\) rather than \(a\).
Practice
Positive numbers \(m\) and \(n\) satisfy \(m^2+n^2=289\) and \(m^2-n^2=161\). The system is linear in \(m^2\) and \(n^2\), so name those two pieces. Enter \(m+n\).
Show the solution
With \(u=m^2\) and \(v=n^2\) the system is \(u+v=289\) and \(u-v=161\). Adding gives \(2u=450\), so \(u=225\) and \(v=64\). Both numbers are positive, so \(m=15\) and \(n=8\), and \(m+n=\boxed{23}\). The naming is what makes the system linear, since \(m^2\) and \(n^2\) appear only as whole pieces.