Every solution below shows the steps and a check in the original equations. Final answers are in bold. If your answer differs, substitute it into both equations to find which one fails, then look for a sign error in the elimination or substitution step.
Problem 1: Is it a solution?
First equation: \(2(2)-1=3\), which matches \(y=3\).
Second equation: \(-2+5=3\), which also matches.
The pair satisfies both equations, so yes, \((2,3)\) is a solution.
Problem 2: Test an ordered pair
First equation: \(4+1=5\), which is true.
Second equation: \(4-1=3\), but the equation needs 2, so it is false.
The pair fails the second equation, so no, \((4,1)\) is not a solution.
Problem 3: Read the intersection
The lines cross at \((2,1)\).
Check: \(2-1=1\) and \(-2(2)+5=1\). Both are true.
The solution is \((2,1)\).
Problem 4: Solve by graphing
The line \(y=x+2\) has y-intercept 2 and slope 1. The line \(y=-x+4\) has y-intercept 4 and slope \(-1\).
They cross at \((1,3)\).
Check: \(1+2=3\) and \(-1+4=3\). The solution is \((1,3)\).
Problem 5: Graph two lines
The line \(y=2x\) passes through \((0,0)\) and \((1,2)\). The line \(y=-x+6\) passes through \((0,6)\) and \((6,0)\).
They cross at \((2,4)\). Check: \(2(2)=4\) and \(-2+6=4\).
The solution is \((2,4)\).
Problem 6: Standard form on a grid
Intercepts of \(x+y=6\): \((6,0)\) and \((0,6)\). Intercepts of \(x-y=2\): \((2,0)\) and \((0,-2)\).
The lines cross at \((4,2)\). Check: \(4+2=6\) and \(4-2=2\).
The solution is \((4,2)\).
Problem 7: Substitute a product
Replace \(y\) with \(3x\): \(x+3x=12\).
\(4x=12\), so \(x=3\).
\(y=3(3)=9\).
Check: \(3+9=12\). The solution is \((3,9)\).
Problem 8: Substitute a sum
Replace \(y\) with \(x+4\): \(2x+(x+4)=13\).
\(3x+4=13\), so \(3x=9\) and \(x=3\).
\(y=3+4=7\).
Check: \(2(3)+7=13\). The solution is \((3,7)\).
Problem 9: Two expressions for y
Both expressions equal \(y\), so set them equal: \(2x-3=x+2\).
Subtract \(x\): \(x-3=2\). Add 3: \(x=5\).
\(y=5+2=7\).
Check: \(2(5)-3=7\). The solution is \((5,7)\).
Problem 10: Substitute for x
Replace \(x\) with \(2y+1\): \(3(2y+1)-y=8\).
Distribute: \(6y+3-y=8\), so \(5y=5\) and \(y=1\).
\(x=2(1)+1=3\).
Check: \(3(3)-1=8\). The solution is \((3,1)\).
Problem 11: Add the equations
Add the equations: \(2x=14\), so \(x=7\).
Then \(7+y=10\) gives \(y=3\).
Check: \(7-3=4\). The solution is \((7,3)\).
Problem 12: Opposite coefficients
The \(y\)-terms are opposites. Add: \(5x=20\), so \(x=4\).
Then \(2(4)+y=11\) gives \(y=3\).
Check: \(3(4)-3=9\). The solution is \((4,3)\).
Problem 13: Eliminate y
Add the equations: \(4x=20\), so \(x=5\).
Then \(5-2y=1\) gives \(-2y=-4\) and \(y=2\).
Check: \(3(5)+2(2)=19\). The solution is \((5,2)\).
Problem 14: Multiply one equation
Multiply the second equation by \(-2\): \(-2x-2y=-10\).
Add to the first equation: \(y=3\).
Then \(x+3=5\) gives \(x=2\).
Check: \(2(2)+3(3)=13\). The solution is \((2,3)\).
Problem 15: Make y cancel
Multiply the second equation by 2: \(4x-2y=2\).
Add to the first: \(7x=14\), so \(x=2\).
Then \(2(2)-y=1\) gives \(y=3\).
Check: \(3(2)+2(3)=12\). The solution is \((2,3)\).
Problem 16: Parallel lines
Both lines have slope 2, but the y-intercepts are 1 and \(-3\).
Parallel lines never cross. Setting the equations equal gives \(2x+1=2x-3\), which simplifies to \(1=-3\), which is false.
The system has no solution.
Problem 17: No solution by inspection
The same expression \(3x+2y\) would have to equal both 5 and 6 for the same \(x\) and \(y\), which is impossible.
Subtracting the equations gives \(0=-1\), a false statement.
No solution; the lines are parallel.
Problem 18: Same line twice
Solve the second equation for \(y\): \(-2y=-6x+4\), so \(y=3x-2\).
This is the same equation as the first, so the graphs are the same line.
The system has infinitely many solutions; every point on the line works.
Problem 19: Movie tickets
Let \(a\) be adult tickets and \(c\) be child tickets.
\(a+c=8\) and \(9a+6c=57\).
Substitute \(c=8-a\): \(9a+6(8-a)=57\), so \(3a+48=57\) and \(a=3\).
\(c=8-3=5\).
Check: \(9(3)+6(5)=27+30=57\). She bought 3 adult and 5 child tickets.
Problem 20: Saving money
Let \(w\) be weeks and \(s\) the savings: \(s=20+5w\) and \(s=50+2w\).
Set equal: \(20+5w=50+2w\), so \(3w=30\) and \(w=10\).
\(s=20+5(10)=70\). Check: \(50+2(10)=70\).
They have equal savings after 10 weeks, with $70 each.
Problem 21: Phone plans
Let \(m\) be minutes: \(C=20+0.10m\) and \(C=10+0.20m\).
Set equal: \(20+0.10m=10+0.20m\), so \(10=0.10m\) and \(m=100\).
Cost: \(20+0.10(100)=30\). Check: \(10+0.20(100)=30\).
The plans cost the same at 100 minutes, $30.
Problem 22: Bake sale
Let \(m\) be muffins and \(k\) cookies: \(m+k=30\) and \(3m+2k=70\).
Multiply the first by \(-2\): \(-2m-2k=-60\). Add to the second: \(m=10\).
\(k=30-10=20\).
Check: \(3(10)+2(20)=70\). He sold 10 muffins and 20 cookies.
Problem 23: Coins in a jar
Let \(n\) be nickels and \(d\) dimes: \(n+d=15\) and \(5n+10d=120\).
Substitute \(n=15-d\): \(5(15-d)+10d=120\), so \(75+5d=120\) and \(d=9\).
\(n=6\).
Check: \(6(5)+9(10)=30+90=120\). She has 6 nickels and 9 dimes.
Problem 24: Two numbers
Let the numbers be \(x\) and \(y\): \(x+y=24\) and \(x-y=6\).
Add: \(2x=30\), so \(x=15\). Then \(y=24-15=9\).
Check: \(15-9=6\). The numbers are 15 and 9.
See every chapter on the Grade 8 math page.