QuanticaAlgebra IOpen in the courseLesson 5.2 closed on a system with no clean isolate, \(9x+7y=76\) and \(5x+8y=72\). Every coefficient is bigger than \(1\), so solving for a letter starts with a division, \(y=\frac{76-9x}{7}\), and fractions appear in every line after that. Set the system aside. It returns as the last problem of this lesson.
Problem
At the solution of the system \(4x+7y=50\) and \(3x-7y=13\), each equation's two sides are the same number, so the sum of the left sides equals the sum of the right sides. Write that sum equation, combine like terms on each side, then solve what remains and enter \(x\).
Show a hint
- Stack the two equations and add straight down, left side to left side and right side to right side.
- The \(+7y\) and \(-7y\) sum to zero, so the sum equation is \(7x=63\).
Show the full solution
Adding the two left sides and the two right sides gives \(7x+0y=63\), so \(7x=63\) and \(x=\boxed{9}\). Back-substituting, \(7y=50-36=14\), so \(y=2\), and \((9,2)\) checks in both originals, \(36+14=50\) and \(27-14=13\). One addition removed \(y\) and left a chapter 4 equation.
Problem
Solve the system \(7x+3y=9\) and \(2x-3y=18\) by adding the equations, then back-substitute into either original. The ask is \(y\), not \(x\).
Show a hint
- The opposite \(y\)-coefficients cancel when the equations are added, but the sum equation has no \(y\) in it, so it cannot answer the ask by itself.
- Adding gives \(9x=27\). Find \(x\), then substitute that value into either original equation.
Show the full solution
Adding gives \(9x=27\), so \(x=3\). Then \(21+3y=9\), \(3y=-12\), and \(y=\boxed{-4}\). Elimination finds one letter and back-substitution finds the other, so stopping at \(x=3\) leaves the ask unanswered.
Problem
The \(y\)-coefficients of \(7x+2y=62\) and \(3x+2y=30\) are equal, not opposite, so adding gives \(10x+4y=92\) and both letters are still there. Subtract the second equation from the first instead, every term on both sides, back-substitute, and enter \(x+y\).
Show a hint
- Subtracting equal numbers from equal numbers is as legal as adding them, and equal coefficients cancel under subtraction.
- The left sides subtract term by term, \(7x-3x\) and \(2y-2y\), and the right sides subtract as well.
Show the full solution
Subtracting gives \(4x+0y=32\), so \(x=8\). Then \(56+2y=62\), \(2y=6\), \(y=3\), and \(x+y=\boxed{11}\). The subtraction has to reach the right sides too, \(62-30=32\), not only the columns of letters.
Problem
Subtracting \(5x-2y=34\) from \(5x+3y=74\), a student writes \(3y-2y=40\), so \(y=40\), and the check fails. Find the sign the student dropped, redo the subtraction with every term carrying its sign, and enter the correct \(y\).
Show a hint
- Subtracting a whole equation subtracts each of its terms, signs included. Look at what the second equation's \(y\)-term actually is.
- That term is \(-2y\), so the \(y\)-column is \(3y-(-2y)\), not \(3y-2y\). Redo it, then the right sides.
Show the full solution
The \(y\)-column is \(3y-(-2y)=5y\) and the right sides give \(74-34=40\), so \(5y=40\) and \(y=\boxed{8}\). Subtracting a whole equation puts a minus in front of parentheses, the rule from 4.4, since the left column means \(5x+3y-(5x-2y)\). The student subtracted \(2y\) instead of \(-2y\).
Matched coefficients are luck. When neither equation offers one, 4.2 allows multiplying an equation, both sides, by any nonzero number, and the result is an equivalent equation with the same solutions. So a matching coefficient can be built. Multiply one equation by whatever number its coefficient needs, then add or subtract.
Problem
In \(3x+4y=59\) and \(6x+5y=85\), no coefficient matches yet. Multiply the first equation, both sides, by \(2\), then subtract to eliminate \(x\). Solve the system and enter \(x+y\).
Show a hint
- Write the doubled first equation out in full, then subtract the second from it column by column, the \(x\) terms, the \(y\) terms, and the right sides.
- The doubled first equation is \(6x+8y=118\). Subtract the second equation from that one.
Show the full solution
Doubling the first equation gives \(6x+8y=118\), and subtracting the second leaves \(3y=33\), so \(y=11\). Then \(3x+44=59\) gives \(x=5\), and \(x+y=\boxed{16}\). The scaling has to reach the right side, so the doubled equation ends in \(118\), not \(59\).
Problem
Solve \(7x+5y=-1\) and \(3x+15y=-39\). Scale the first equation by \(3\), subtract the second, and enter \(y\). Both right sides are negative, so the subtraction there needs care.
Show a hint
- Tripling the first equation makes both \(y\)-columns \(15y\). Scale both sides, so the \(-1\) is tripled as well.
- The right sides subtract as \(-3-(-39)\), a difference of two negatives.
Show the full solution
Tripling the first equation gives \(21x+15y=-3\), and subtracting the second leaves \(18x=-3-(-39)=36\), so \(x=2\). Then \(14+5y=-1\) gives \(5y=-15\) and \(y=\boxed{-3}\). The tripled right side is \(-3\), not \(-1\), and subtracting \(-39\) adds \(39\).
Problem
In \(6x+7y=56\) and \(3x+5y=31\), one letter eliminates by scaling a single equation, while the other needs both equations scaled to reach \(35\). Pick the cheap one, solve the system, and enter the product \(xy\).
Show a hint
- Price the two choices before scaling anything. The \(x\)-coefficients are \(6\) and \(3\), the \(y\)-coefficients are \(7\) and \(5\).
- Doubling the second equation makes both \(x\)-columns \(6x\). Subtract, and a \(y\)-equation is left.
Show the full solution
Doubling the second equation gives \(6x+10y=62\), and subtracting the first leaves \(3y=6\), so \(y=2\). Then \(6x+14=56\) gives \(x=7\), and \(xy=\boxed{14}\). Eliminating \(y\) reaches the same pair, but it means scaling to \(35y\) and subtracting \(280-217\) instead of \(62-56\).
Problem
Eliminate \(y\) from \(5x+4y=46\) and \(7x-6y=-40\) by scaling both equations so the \(y\)-terms become \(12y\) and \(-12y\). Solve the system, check the pair in both originals, and enter the product \(xy\).
Show a hint
- Each equation gets multiplied by whatever its \(y\)-coefficient needs to reach \(12\), and the factor hits both sides, right side included.
- The scaled \(y\)-terms are \(12y\) and \(-12y\), which are opposites, so the two equations are added rather than subtracted.
Show the full solution
Three times the first is \(15x+12y=138\), twice the second is \(14x-12y=-80\), and adding them gives \(29x=58\), so \(x=2\). Then \(10+4y=46\) gives \(y=9\), and \(xy=\boxed{18}\). Check \((2,9)\) in both originals, \(10+36=46\) and \(14-54=-40\).
Problem
Solve \(4x+9y=32\) and \(2x+7y=21\) by eliminating \(x\). Every line stays whole until one final division. Enter \(x\) as a fraction in lowest terms.
Show a hint
- Doubling the second equation matches the \(x\)-columns at \(4x\).
- \(y\) comes out whole. Put that \(y\) back into either equation and the coefficient of \(x\) will not divide the right side evenly, which is where the fraction finally shows up.
Show the full solution
Doubling the second equation gives \(4x+14y=42\), and subtracting the first leaves \(5y=10\), so \(y=2\). Then \(2x+14=21\) gives \(2x=7\) and \(x=\boxed{\frac{7}{2}}\). Every line before that division stayed whole, so the only fraction in the problem is the answer itself.
Both letters can cancel at once. When they do, what remains is a statement about numbers alone, true or false on its own. Substitution reached the same situation in 5.2, and the reading is the same here. The leftover statement is a verdict on the whole system, and the next two problems have one of each kind.
Problem
Try elimination on \(2x+6y=35\) and \(3x+9y=48\). Scale the first equation by \(3\) and the second by \(2\), then subtract. Both letters cancel and a statement about numbers is left. Decide what it means and enter the number of solutions of the system.
Show a hint
- When both letter columns cancel, the number statement left behind is the verdict on the whole system.
- Both scalings give the left side \(6x+18y\), while the right sides come out \(105\) and \(96\). Subtract and read what is left.
Show the full solution
The scaled equations are \(6x+18y=105\) and \(6x+18y=96\), and subtracting gives \(0=9\), false at every pair, so the system has \(\boxed{0}\) solutions. That is 5.1's count for conflicting claims, reached here by elimination and in 5.2 by substitution.
Problem
Eliminating on \(3x-5y=12\) and \(6x-10y=24\) leaves \(0=0\), true at every pair, so the system has infinitely many solutions. That does not make every pair a solution. Of \((4,0)\), \((19,9)\), \((14,6)\), \((8,2)\), enter how many solve the system.
Show a hint
- \(0=0\) means the second equation is a scaled copy of the first, so a pair solves the system exactly when it solves \(3x-5y=12\).
- Substitute each listed pair into \(3x-5y=12\) and count the ones that give \(12\).
Show the full solution
In \(3x-5y\) the four pairs give \(12\), \(12\), \(12\), and \(24-10=14\), so \(\boxed{3}\) of them solve the system. Infinitely many solutions means every pair satisfying the shared equation, not every pair anybody writes down.
Back to the shelved system. In \(9x+7y=76\) and \(5x+8y=72\) no letter isolates without a division, which is what stopped substitution. Elimination isolates nothing, so no division is needed. The \(x\)-coefficients \(9\) and \(5\) both divide \(45\), so scaling the first equation by \(5\) and the second by \(9\) puts \(45x\) in each.
Problem
Finish the system 5.2 left unsolved. Scale \(9x+7y=76\) by \(5\) and \(5x+8y=72\) by \(9\), so both carry \(45x\), then subtract and solve. Enter \(y\) as a fraction in lowest terms.
Show a hint
- Multiplying an equation through by a number keeps the same solution pair, so scaling is safe. The target \(45\) is the smallest number that both \(9\) and \(5\) divide into.
- Five times the first equation is \(45x+35y=380\) and nine times the second is \(45x+72y=648\). One subtraction and one division are left.
Show the full solution
The scaled equations are \(45x+35y=380\) and \(45x+72y=648\), and subtracting the first from the second leaves \(37y=268\), so \(y=\boxed{\frac{268}{37}}\), with \(x=\frac{104}{37}\). The work stays in whole numbers until that last division, while isolating a letter in either original equation would start with a division.
Both methods solve every linear system in two variables, so the choice is convenience. An equation already solved for a letter, or a letter with coefficient \(1\) or \(-1\), points to substitution. Coefficients that match, or match after one scaling, point to elimination. Lesson 5.4 takes on systems that need cleanup before either method applies.
Practice these ideas
Practice
The \(y\)-terms of \(6x+5y=57\) and \(3x-5y=6\) are \(+5y\) and \(-5y\). Add the left sides and add the right sides, solve the one-letter equation that remains, and enter \(x\).
Show the solution
Adding gives \(9x=63\), so \(x=\boxed{7}\), and \(5y=57-42=15\) gives \(y=3\). Check in the second equation, \(21-15=6\).
Practice
Solve \(6x+y=40\) and \(3x-y=14\) by adding the two equations, then back-substituting into either original. Enter \(y\).
Show the solution
Adding gives \(9x=54\), so \(x=6\), and \(6(6)+y=40\) gives \(y=\boxed{4}\). Check in the second equation, \(3(6)-4=14\). Adding removes one letter and back-substitution finds the other.
Practice
The \(x\)-terms of \(-4x+9y=42\) and \(4x+11y=78\) are already opposites, so adding removes \(x\) and leaves an equation in \(y\). Solve the system and enter \(x\).
Show the solution
Adding the equations gives \(20y=120\), so \(y=6\). The first equation becomes \(-4x+54=42\), so \(-4x=-12\) and \(x=\boxed{3}\). The second equation checks, \(4(3)+11(6)=12+66=78\).
Practice
Two numbers add to \(94\) and differ by \(22\). Write the two facts as a system, add the equations to eliminate a letter, and enter the smaller number.
Show the solution
The system is \(x+y=94\) and \(x-y=22\). Adding gives \(2x=116\), so \(x=58\), and \(58+y=94\) gives \(y=\boxed{36}\). Check, \(58-36=22\). A sum fact and a difference fact always add straight to a one-letter equation.
Practice
The \(y\)-coefficients of \(9x+5y=120\) and \(4x+5y=70\) are both \(5\), equal rather than opposite, so adding keeps both letters. Subtract the second equation from the first instead, every term on both sides, and enter \(x\).
Show the solution
Subtracting the second equation from the first gives \(5x=50\), so \(x=\boxed{10}\). Then \(5y=70-40=30\) gives \(y=6\), and the first equation checks, \(90+30=120\). Subtracting the other way gives \(-5x=-50\) and the same \(x\).
Practice
Subtract \(2x-3y=-11\) from \(2x+6y=52\). The \(x\)-column cancels. Finish the solve and enter \(x\).
Show the solution
Subtracting gives \(9y=63\), so \(y=7\), and \(2x+42=52\) gives \(2x=10\) and \(x=\boxed{5}\). The pair checks in \(2x-3y=-11\), since \(10-21=-11\). Both the \(-3y\) and the \(-11\) flipped sign when that equation was subtracted.
Practice
Both \(7x+5y=81\) and \(7x-3y=-15\) lead with \(7x\). Subtract the second equation from the first, flipping the sign of every term of the subtracted equation, the right side included. Enter \(y\).
Show the solution
Subtracting term by term, \(5y-(-3y)=8y\) and \(81-(-15)=96\), so \(8y=96\) and \(y=\boxed{12}\). Back-substituting gives \(x=3\), and \(7(3)-3(12)=21-36=-15\) checks the second equation. Writing \(5y-3y\) instead of \(5y+3y\) is the usual slip.
Practice
In \(8x+5y=73\) and \(4x+15y=99\), scaling the first equation once makes the \(y\)-coefficients match. Make that match, eliminate \(y\), and enter \(x\).
Show the solution
Three times the first equation is \(24x+15y=219\), and subtracting the second gives \(20x=120\), so \(x=\boxed{6}\). Then \(48+5y=73\) gives \(y=5\), and \(4(6)+15(5)=99\) checks. Scaling has to reach the right side too.
Practice
Solve \(5x+9y=87\) and \(10x+7y=86\). Doubling the first equation, both sides, lines up the \(x\)-terms at \(10x\). Enter \(y\).
Show the solution
Twice the first equation is \(10x+18y=174\). Subtracting the second gives \(11y=88\), so \(y=\boxed{8}\), with \(x=3\) from \(5x+72=87\). Check in the second equation, \(30+56=86\).
Practice
No coefficients in \(9x+y=43\) and \(5x+3y=19\) match yet, but one scaling of the first equation, both sides, brings the \(y\)-columns together. Solve the system and enter \(y\).
Show the solution
Three times the first equation is \(27x+3y=129\). Subtracting the second gives \(22x=110\), so \(x=5\), and \(45+y=43\) gives \(y=\boxed{-2}\). Check in the second equation, \(25-6=19\).
Practice
For \(3x+7y=41\) and \(12x+9y=69\), the \(x\)-coefficients first meet at \(12\) and the \(y\)-coefficients first meet at \(63\). Enter the letter that is cheaper to eliminate.
Show the solution
Matching at \(12x\) takes one scaling, the first equation times \(4\), while matching at \(63y\) takes two, the first times \(9\) and the second times \(7\), so the cheaper letter is \(\boxed{x}\). Both routes end at the same pair \((2,5)\), and the only difference is how large the numbers get.
Practice
Solve \(6x+7y=75\) and \(4x+5y=51\). Neither \(x\)-coefficient divides the other, so scale both equations to eliminate \(x\). Enter the product \(xy\).
Show the solution
Twice the first is \(12x+14y=150\) and three times the second is \(12x+15y=153\). Subtracting gives \(y=3\), and \(6x+21=75\) gives \(6x=54\) and \(x=9\), so \(xy=\boxed{27}\). Check in the second equation, \(36+15=51\).
Practice
The \(x\)-coefficients of \(2x+3y=16\) and \(7x+5y=23\) are \(2\) and \(7\), whose only common factor is \(1\). Scale both equations to a common \(x\)-coefficient, subtract, and enter \(x\).
Show the solution
Seven times the first is \(14x+21y=112\) and twice the second is \(14x+10y=46\). Subtracting gives \(11y=66\), so \(y=6\), and \(2x+18=16\) gives \(2x=-2\) and \(x=\boxed{-1}\). Check in the second equation, \(-7+30=23\). A solution pair can have a negative member.
Practice
Solve \(7x+9y=29\) and \(6x+3y=17\). Nothing here is fractional until the last step. Enter \(y\) as a fraction in lowest terms.
Show the solution
Three times the second equation is \(18x+9y=51\), and subtracting the first gives \(11x=22\), so \(x=2\). Then \(6(2)+3y=17\) gives \(3y=5\) and \(y=\boxed{\frac{5}{3}}\). Check in the first equation, \(7(2)+9\cdot\frac{5}{3}=14+15=29\).
Practice
Eliminate a letter from \(3x+8y=44\) and \(6x+16y=77\). Both letters cancel at once and a number statement is left. Read it and enter the number of solutions of the system.
Show the solution
Twice the first equation is \(6x+16y=88\). Subtracting the second gives \(0=11\), false at every pair, so the system has \(\boxed{0}\) solutions. The same left side cannot be \(88\) and \(77\) at once, so the two equations contradict each other and no pair passes both.
Practice
Elimination on \(5x-2y=15\) and \(15x-6y=45\) leaves \(0=0\), true everywhere, so the system has infinitely many solutions. That does not make every pair one of them. Of \((3,0)\), \((5,5)\), \((6,7)\), \((4,1)\), enter how many solve the system.
Show the solution
The left sides are \(15-0=15\), \(25-10=15\), \(30-14=16\), and \(20-2=18\), so \(\boxed{2}\) of the four solve the system. Infinitely many solutions means every pair satisfying the shared equation, and only those.
Practice
For exactly one value of \(k\), elimination on \(8x+12y=372\) and \(2x+3y=k\) leaves \(0=0\) and the system has infinitely many solutions. Every other value of \(k\) leaves a false statement instead. Enter \(k\).
Show the solution
Since \(8x+12y\) is \(4\) times \(2x+3y\), the two claims agree at every pair exactly when \(372=4k\), so \(k=\boxed{93}\). Check, multiplying \(2x+3y=93\) by \(4\) gives \(8x+12y=372\). Any other value makes one left side equal two different numbers, and the count drops to zero.
Practice
Solve \(10x+6y=162\) and \(7x+9y=147\). One letter's coefficients meet at \(18\) and the other's meet at \(70\). Eliminate the cheaper one, scaling both equations, and enter the product \(xy\).
Show the solution
Three times the first is \(30x+18y=486\) and twice the second is \(14x+18y=294\). Subtracting gives \(16x=192\), so \(x=12\), and \(120+6y=162\) gives \(6y=42\) and \(y=7\), so \(xy=\boxed{84}\). Check in the second equation, \(84+63=147\). Eliminating \(x\) instead would push the scaled equations past \(1000\).