QuanticaAlgebra IOpen in the courseTwo numbers satisfy \(y=3x+26\) and \(x+y=2\). Run 5.1's search on the second equation, \((0,2)\), \((1,1)\), \((2,0)\), and every candidate fails the first. The pair hides among negatives, where listing never ends. The first equation is not one more claim to test. It says what \(y\) is. At the solution, \(y\) and \(3x+26\) are the same number.
Problem
The first equation says \(y\) equals \(3x+26\), so write \(3x+26\) in \(y\)'s place in \(x+y=2\). One equation in one variable remains. Solve it and enter \(x\).
Show a hint
- The swap leaves an equation with only \(x\) in it, exactly the kind chapter 4 solves.
- After the substitution the claim reads \(x+(3x+26)=2\). Collect the left side to \(4x+26=2\).
Show the full solution
\(x+(3x+26)=2\) collects to \(4x+26=2\), so \(4x=-24\) and \(x=\boxed{-6}\). The first equation names \(y\), so its expression can stand wherever \(y\) stands.
Problem
The partner is still missing. Substitute \(x=-6\) into \(y=3x+26\), the equation already solved for \(y\), and enter \(y\). Check the pair in both original equations first.
Show a hint
- The solved-form equation computes \(y\) from any \(x\), including this one.
- Substituting gives \(y=3(-6)+26\). Keep the parentheses around the negative value.
Show the full solution
\(y=3(-6)+26=-18+26=\boxed{8}\). Check \((-6,8)\) in both originals, \(8=3(-6)+26\) and \(-6+8=2\), so the pair passes the full 5.1 test.
Problem
The system \(x=3y+22\) and \(x+2y=112\) leads with \(x\) this time. Replace \(x\) in the second equation with \(3y+22\), solve for \(y\), back-substitute, and enter \(x\).
Show a hint
- Either letter can be the pre-solved one. The steps do not change. Replace the letter that equals an expression.
- \((3y+22)+2y=112\) collects to \(5y+22=112\).
Show the full solution
\((3y+22)+2y=112\) gives \(5y+22=112\), so \(5y=90\), \(y=18\), and \(x=3(18)+22=\boxed{76}\). The ask is \(x\), so the back-substitution is not optional.
So far the replaced letter stood alone. In \(4x+3y=47\) the \(y\) is tripled, and \(3y\) means three copies of all of \(y\), so whatever stands in for \(y\) must arrive as one unit. With \(y=2x+7\), the incoming expression takes the seat as a block, not a piece at a time.
Problem
Solve the system \(y=3x-16\) and \(5x+2y=78\). The \(2y\) doubles the whole expression, so substitute in parentheses, distribute, and enter \(x+y\).
Show a hint
- Substitute the whole expression \(3x-16\) for \(y\) in the second equation, with parentheses around all of it.
- \(5x+2(3x-16)=78\) distributes to \(11x-32=78\).
Show the full solution
\(5x+2(3x-16)=78\) gives \(11x-32=78\), so \(11x=110\) and \(x=10\), then \(y=3(10)-16=14\) and \(x+y=\boxed{24}\).
Problem
A student substitutes \(y=2x-3\) into \(x+4y=51\) and writes \(x+4\cdot 2x-3=51\), no parentheses. That line gives \(x=6\), then \(y=9\), and the check fails, since \(6+4(9)=42\), not \(51\). Redo the substitution with the parentheses it needed and enter the correct \(x\).
Show a hint
- In \(4y\), the \(4\) must multiply everything \(y\) is, not just the first piece.
- The faithful line is \(x+4(2x-3)=51\). Distribute, then collect.
Show the full solution
\(x+4(2x-3)=51\) gives \(9x-12=51\), so \(9x=63\) and \(x=\boxed{7}\), with \(y=2(7)-3=11\) and \(7+44=51\). The bare line subtracted \(3\) where the equation subtracts \(12\).
Nobody pre-solves equations for you outside a lesson. When neither equation arrives in solved form, pick a letter, solve for it with 4.3's moves, then substitute exactly as before. One isolate is the whole cost, and the rest of the run does not change.
Problem
Neither equation of \(3x+y=32\) and \(7x+2y=75\) is solved for a letter, but the first is one 4.3 move away. Solve it for \(y\), substitute into the second, and enter \(y\). Check the pair in both originals.
Show a hint
- Subtracting \(3x\) from both sides pre-solves the first equation for \(y\).
- Substitute \(y=32-3x\) into the second equation to get \(7x+2(32-3x)=75\).
Show the full solution
\(y=32-3x\), so \(7x+2(32-3x)=75\) collects to \(x+64=75\) and \(x=11\), then \(y=32-33=\boxed{-1}\). Check, \(33+(-1)=32\) and \(77-2=75\), both originals pass.
Problem
In \(3x-y=20\) and \(5x+2y=4\) the \(y\) has coefficient \(-1\). Solving the first for \(y\) still takes one move, \(y=3x-20\). Substitute into the second equation and enter \(y\).
Show a hint
- Coefficient \(-1\) is as cheap as coefficient \(1\). No dividing happens on the way to solved form.
- \(5x+2(3x-20)=4\) collects to \(11x-40=4\).
Show the full solution
\(5x+2(3x-20)=4\) gives \(11x-40=4\), so \(11x=44\), \(x=4\), and \(y=3(4)-20=\boxed{-8}\). Nothing stops a solution pair from being negative.
Problem
One letter in \(4x+y=198\) and \(2x+3y=204\) isolates without fractions. Find it, isolate it, substitute, solve the system, and enter the product \(xy\).
Show a hint
- Only \(y\) in the first equation has coefficient \(1\), so isolating it costs one 4.3 move and no fractions.
- Substituting \(y=198-4x\) into the second equation gives \(2x+3(198-4x)=204\).
Show the full solution
Isolate \(y=198-4x\), then \(2x+3(198-4x)=204\) gives \(2x+594-12x=204\), so \(-10x=-390\) and \(x=39\), then \(y=198-156=42\). The product is \(39\cdot 42=\boxed{1638}\). Check, \(2(39)+3(42)=78+126=204\).
Problem
Solve the system \(y=2x+4\) and \(6x+2y=33\). Nothing requires the pair to be whole. Check the pair in both equations, then enter \(x\) as a fraction in lowest terms.
Show a hint
- The run is the usual one. Only the ending is not a whole number.
- \(6x+2(2x+4)=33\) collects to \(10x+8=33\).
Show the full solution
\(6x+2(2x+4)=33\) gives \(10x+8=33\), so \(10x=25\) and \(x=\frac{25}{10}=\boxed{\frac{5}{2}}\), with \(y=2\cdot\frac{5}{2}+4=9\). Check, \(9=5+4\) and \(15+18=33\), both originals pass.
Sometimes the substitution erases the letter entirely. Every \(x\)-term cancels, and a statement with no variable in it is left behind, true or false on its own. 4.3 met this inside one equation. Here the leftover statement is a verdict about the whole system, and the next two problems read one of each kind.
Problem
Substitute \(y=4x+5\) into \(8x-2y=6\) and simplify. Every \(x\)-term cancels and a number statement is left. Decide what it means and enter the number of solutions of the system.
Show a hint
- 4.3 met a vanishing variable inside one equation. The number statement left behind is the verdict.
- \(8x-2(4x+5)\) simplifies to \(-10\), and the equation claims that equals \(6\).
Show the full solution
\(8x-2(4x+5)=8x-8x-10=-10\), so the substituted equation reads \(-10=6\), false at every \(x\), and the system has \(\boxed{0}\) solutions. No pair can make both claims true at once.
Problem
Substituting \(y=2x+6\) into \(4x-2y=-12\) leaves \(-12=-12\), true for every \(x\). Infinitely many pairs solve the system, but not every pair does. Of \((2,10)\), \((6,18)\), \((10,26)\), \((8,20)\), enter how many solve the system.
Show a hint
- Every pair solving the first equation solves both. Pairs off it solve neither.
- Test each pair in \(y=2x+6\), first value in for \(x\), second in for \(y\).
Show the full solution
\(10=4+6\), \(18=12+6\), and \(26=20+6\) pass, but \(2(8)+6=22\neq 20\), so \(\boxed{3}\) of the four solve the system. Infinitely many means every pair satisfying the equation, and only those.
Problem
Two numbers satisfy \(x-2y=45\) and \(2x+3y=398\). Isolate the coefficient-1 letter, substitute, solve the system, check the pair in both equations, and enter the product \(xy\).
Show a hint
- Only \(x\) carries coefficient \(1\). One 4.3 move gives \(x=2y+45\).
- Substituting gives \(2(2y+45)+3y=398\), which collects to \(7y+90=398\).
Show the full solution
Isolate \(x=2y+45\), then \(2(2y+45)+3y=398\) gives \(7y+90=398\), so \(7y=308\) and \(y=44\), then \(x=2(44)+45=133\). Check, \(133-88=45\) and \(266+132=398\). The product is \(133\cdot 44=\boxed{5852}\).
All the systems in this lesson had a letter with coefficient \(1\) or \(-1\), or were already solved for a letter. There is no letter to isolate cleanly in \(9x+7y=76\) and \(5x+8y=72\), so any isolation starts with a division and brings fractions into every later step. Lesson 5.3 introduces elimination, a second method for exactly this case.
Practice these ideas
Practice
The numbers \(x\) and \(y\) satisfy \(y=x+25\) and \(x+y=71\). Substitute the expression for \(y\) into the second equation, solve, and enter \(x\).
Show the solution
Substituting gives \(x+(x+25)=71\), so \(2x+25=71\), \(2x=46\), and \(x=\boxed{23}\). Then \(y=48\), and \(23+48=71\) checks.
Practice
The system \(y=2x\) and \(3x+y=40\). The expression for \(y\) has no constant term, and the steps are exactly the same. Substitute, solve, back-substitute, and enter \(y\).
Show the solution
Substituting gives \(3x+2x=40\), so \(5x=40\) and \(x=8\). Back-substituting, \(y=2(8)=\boxed{16}\). The pair checks in the second equation, since \(24+16=40\).
Practice
The system \(x=y+10\) and \(x+4y=30\) leads with \(x\). Replace the \(x\) of the second equation with \(y+10\), solve for \(y\), back-substitute, and enter \(x\).
Show the solution
\((y+10)+4y=30\) collects to \(5y+10=30\), so \(5y=20\) and \(y=4\). Then \(x=4+10=\boxed{14}\). Check, \(14+4(4)=30\).
Practice
Solve the system \(y=2x+34\) and \(2x+y=150\). Substitute, solve for \(x\), then finish with a back-substitution and enter \(y\).
Show the solution
\(2x+(2x+34)=150\) gives \(4x=116\), so \(x=29\), and \(y=2(29)+34=\boxed{92}\). Check, \(58+92=150\).
Practice
In \(y=x-5\) and \(3x+2y=50\), the \(y\) slot carries a coefficient, so the substitution needs parentheses. Substitute, distribute, solve the system, and enter \(x+y\).
Show the solution
\(3x+2(x-5)=50\) distributes to \(3x+2x-10=50\), so \(5x=60\) and \(x=12\). Then \(y=12-5=7\) and \(x+y=\boxed{19}\). Check, \(3(12)+2(7)=36+14=50\). Dropping the parentheses would give \(3x+2x-5=50\), which subtracts \(5\) where the equation subtracts \(10\).
Practice
Solve the system \(y=2x+1\) and \(5x+3y=80\). The \(3y\) takes the whole expression as one unit. Substitute with parentheses, distribute, and enter \(y\).
Show the solution
\(5x+3(2x+1)=80\) distributes to \(5x+6x+3=80\), so \(11x=77\) and \(x=7\). Then \(y=2(7)+1=\boxed{15}\), and \(35+45=80\) checks.
Practice
Substituting \(y=3x-7\) into \(2x+5y=61\), a student writes \(2x+5\cdot 3x-7\), no parentheses. At every value of \(x\), the student's expression exceeds the faithful \(2x+5(3x-7)\) by the same fixed amount. Enter that amount.
Show the solution
The faithful expression is \(2x+15x-35=17x-35\) and the bare one is \(2x+15x-7=17x-7\), a gap of \(35-7=\boxed{28}\) at every \(x\). The missing parentheses shrink the subtraction from \(35\) to \(7\).
Practice
Neither equation of \(4x+y=231\) and \(2x+3y=233\) is pre-solved. Isolate the coefficient-1 letter with one 4.3 move, substitute into the other equation, and enter \(x\).
Show the solution
Isolate \(y=231-4x\), then \(2x+3(231-4x)=233\) gives \(2x+693-12x=233\), so \(-10x=-460\) and \(x=\boxed{46}\). Then \(y=231-184=47\), and \(92+141=233\) checks.
Practice
For \(x+5y=14\) and \(3x+4y=64\), the letter that isolates cleanly is \(x\). Isolate it, substitute into the second equation, solve, and enter \(y\).
Show the solution
\(x=14-5y\), so \(3(14-5y)+4y=64\) gives \(42-15y+4y=64\), then \(42-11y=64\), \(-11y=22\), and \(y=\boxed{-2}\), with \(x=14+10=24\). Check, \(24-10=14\) and \(72-8=64\).
Practice
Solve \(x=2y+3\) and \(3x+2y=11\). The system leads with \(x\), and the answer is not a whole number. Enter \(y\) as a fraction in lowest terms.
Show the solution
\(3(2y+3)+2y=11\) gives \(6y+9+2y=11\), so \(8y=2\) and \(y=\frac{2}{8}=\boxed{\frac{1}{4}}\), with \(x=2\cdot\frac{1}{4}+3=\frac{7}{2}\). Check, \(3\cdot\frac{7}{2}+\frac{1}{2}=11\).
Practice
In \(4x+y=248\) and \(2x+3y=254\), one letter isolates without fractions and one does not. Pick well, solve the system, and enter the product \(xy\).
Show the solution
Isolate \(y=248-4x\), then \(2x+3(248-4x)=254\) gives \(2x+744-12x=254\), so \(-10x=-490\) and \(x=49\), then \(y=248-196=52\). The product is \(49\cdot 52=\boxed{2548}\). Check, \(98+156=254\).
Practice
Substitute \(y=5x-3\) into \(10x-2y=12\) and simplify. The letter vanishes, and the left side sits the same fixed amount below \(12\) at every \(x\). Enter that amount.
Show the solution
\(10x-2(5x-3)=10x-10x+6=6\), so the substituted equation reads \(6=12\), and the left side sits \(12-6=\boxed{6}\) below the right side at every \(x\). The statement is false everywhere, so the system has no solution.
Practice
Substituting \(y=4x+1\) into \(12x-3y=-3\) leaves \(-3=-3\), a true statement. Every pair solving the first equation solves the system. Of the pairs \((3,13)\), \((0,1)\), \((2,8)\), \((5,20)\), enter how many solve it.
Show the solution
Testing each pair in \(y=4x+1\), \(13=12+1\) and \(1=0+1\) pass, while \(4(2)+1=9\neq 8\) and \(4(5)+1=21\neq 20\) fail. So \(\boxed{2}\) of the four solve the system.
Practice
For \(x-2y=1\) and \(6x+5y=74\), isolate \(x\), substitute into the second equation, solve the system, and enter \(x+y\).
Show the solution
\(x=2y+1\), so \(6(2y+1)+5y=74\) gives \(12y+6+5y=74\), then \(17y=68\) and \(y=4\), with \(x=2(4)+1=9\). So \(x+y=\boxed{13}\). Check, \(9-8=1\) and \(54+20=74\).
Practice
One number is 5 less than double another, and the two add to 43. In symbols, \(y=2x-5\) and \(x+y=43\). Solve the system and enter the larger number.
Show the solution
\(x+(2x-5)=43\) gives \(3x-5=43\), so \(3x=48\) and \(x=16\), then \(y=2(16)-5=27\). The larger number is \(\boxed{27}\). The problem asks for the larger of the two numbers, not for \(x\), so find both values before answering.
Practice
For \(6x+y=37\) and \(4x+3y=90\), isolate the coefficient-1 letter, substitute, and enter \(x\) as a fraction in lowest terms.
Show the solution
\(y=37-6x\), so \(4x+3(37-6x)=90\) gives \(4x+111-18x=90\), then \(111-14x=90\), \(14x=21\), and \(x=\frac{21}{14}=\boxed{\frac{3}{2}}\), with \(y=37-9=28\). The clean isolate keeps fractions out of the middle of the work, not out of the answer.
Practice
In \(3x-y=106\) and \(2x+2y=220\), the \(y\) has coefficient \(-1\), and 4.3's moves still isolate it, \(y=3x-106\). Substitute, solve the system, and enter the product \(xy\).
Show the solution
Substitute \(y=3x-106\), so \(2x+2(3x-106)=220\) gives \(8x-212=220\), then \(8x=432\) and \(x=54\), with \(y=3(54)-106=56\). The product is \(54\cdot 56=\boxed{3024}\). Check, \(108+112=220\).