Solving systems by substitution involves replacing one variable with an equivalent expression from the other equation. This method works especially well when one equation is already solved for a variable or when a variable can be isolated easily. The problems below include systems with one solution, no solution, infinitely many solutions, fractions, decimals, and real-world applications.
Solve Using Substitution
\(\textbf{1)}\) \(y=-2x+5\)
\(y=x-1\)
The answer is \((2,1)\)
Set the two expressions for \(y\) equal to each other.
\(\,\,\,\,\,-2x+5=x-1\)
Add \(2x\) to both sides.
\(\,\,\,\,\,5=3x-1\)
Add 1 to both sides.
\(\,\,\,\,\,6=3x\)
Divide both sides by 3.
\(\,\,\,\,\,x=2\)
Substitute \(x=2\) into \(y=x-1\).
\(\,\,\,\,\,y=2-1\)
\(\,\,\,\,\,y=1\)
Therefore, the solution is \((2,1)\).
\(\textbf{2)}\) \(y=x+5\)
\(y=2x+4\)
The answer is \((1,6)\)
Set the two expressions for \(y\) equal to each other.
\(\,\,\,\,\,x+5=2x+4\)
Subtract \(x\) from both sides.
\(\,\,\,\,\,5=x+4\)
Subtract 4 from both sides.
\(\,\,\,\,\,x=1\)
Substitute \(x=1\) into \(y=x+5\).
\(\,\,\,\,\,y=1+5\)
\(\,\,\,\,\,y=6\)
Therefore, the solution is \((1,6)\).
\(\textbf{3)}\) \(4x+3y=12\)
\(y=4\)
The answer is \((0,4)\)
Substitute \(y=4\) into the first equation.
\(\,\,\,\,\,4x+3(4)=12\)
Simplify.
\(\,\,\,\,\,4x+12=12\)
Subtract 12 from both sides.
\(\,\,\,\,\,4x=0\)
Divide both sides by 4.
\(\,\,\,\,\,x=0\)
The value of \(y\) is already given as 4.
Therefore, the solution is \((0,4)\).
\(\textbf{4)}\) \(2x-3y=0\)
\(y=x-1\)
The answer is \((3,2)\)
Substitute \(y=x-1\) into the first equation.
\(\,\,\,\,\,2x-3(x-1)=0\)
Distribute the \(-3\).
\(\,\,\,\,\,2x-3x+3=0\)
Combine like terms.
\(\,\,\,\,\,-x+3=0\)
Subtract 3 from both sides.
\(\,\,\,\,\,-x=-3\)
Divide both sides by \(-1\).
\(\,\,\,\,\,x=3\)
Substitute \(x=3\) into \(y=x-1\).
\(\,\,\,\,\,y=3-1\)
\(\,\,\,\,\,y=2\)
Therefore, the solution is \((3,2)\).
\(\textbf{5)}\) \(3x+y=10\)
\(x-y=2\)
The answer is \((3,1)\)
Solve the second equation for \(x\).
\(\,\,\,\,\,x-y=2\)
\(\,\,\,\,\,x=y+2\)
Substitute \(x=y+2\) into the first equation.
\(\,\,\,\,\,3(y+2)+y=10\)
Distribute the 3.
\(\,\,\,\,\,3y+6+y=10\)
Combine like terms.
\(\,\,\,\,\,4y+6=10\)
Subtract 6 from both sides.
\(\,\,\,\,\,4y=4\)
Divide both sides by 4.
\(\,\,\,\,\,y=1\)
Substitute \(y=1\) into \(x=y+2\).
\(\,\,\,\,\,x=1+2\)
\(\,\,\,\,\,x=3\)
Therefore, the solution is \((3,1)\).
\(\textbf{6)}\) \(y=3x+2\)
\(2y-6x=-6\)
The answer is No Solution or Inconsistent.
Substitute \(y=3x+2\) into the second equation.
\(\,\,\,\,\,2(3x+2)-6x=-6\)
Distribute the 2.
\(\,\,\,\,\,6x+4-6x=-6\)
Combine like terms.
\(\,\,\,\,\,4=-6\)
This statement is false, so the system is inconsistent.
The lines are parallel and never intersect.
Therefore, the system has no solution.
\(\textbf{7)}\) \(y=-3x+5\)
\(9x+3y=15\)
The answer is Infinitely Many Solutions or Consistent and Dependent.
Substitute \(y=-3x+5\) into the second equation.
\(\,\,\,\,\,9x+3(-3x+5)=15\)
Distribute the 3.
\(\,\,\,\,\,9x-9x+15=15\)
Combine like terms.
\(\,\,\,\,\,15=15\)
This statement is always true.
The equations represent the same line.
Therefore, the system has infinitely many solutions.
\(\textbf{8)}\) Four times one number added to another number is 20. The second number is 5 more than the first. Find the numbers.
The first number is \(3\). The second number is \(8\).
Let the first number be \(x\) and the second number be \(y\).
Write an equation for the first statement.
\(\,\,\,\,\,4x+y=20\)
Write an equation for the second statement.
\(\,\,\,\,\,y=x+5\)
Substitute \(y=x+5\) into the first equation.
\(\,\,\,\,\,4x+(x+5)=20\)
Combine like terms.
\(\,\,\,\,\,5x+5=20\)
Subtract 5 from both sides.
\(\,\,\,\,\,5x=15\)
Divide both sides by 5.
\(\,\,\,\,\,x=3\)
Substitute \(x=3\) into \(y=x+5\).
\(\,\,\,\,\,y=3+5\)
\(\,\,\,\,\,y=8\)
Therefore, the two numbers are \(3\) and \(8\).
\(\textbf{9)}\) \(y=4x-7\)
\(2x+y=5\)
The answer is \((2,1)\).
Substitute \(y=4x-7\) into the second equation.
\(\,\,\,\,\,2x+(4x-7)=5\)
Combine like terms.
\(\,\,\,\,\,6x-7=5\)
Add 7 to both sides.
\(\,\,\,\,\,6x=12\)
Divide both sides by 6.
\(\,\,\,\,\,x=2\)
Substitute \(x=2\) into \(y=4x-7\).
\(\,\,\,\,\,y=4(2)-7\)
\(\,\,\,\,\,y=8-7\)
\(\,\,\,\,\,y=1\)
Therefore, the solution is \((2,1)\).
\(\textbf{10)}\) \(x=2y+1\)
\(3x-y=13\)
The answer is \((5,2)\).
Substitute \(x=2y+1\) into the second equation.
\(\,\,\,\,\,3(2y+1)-y=13\)
Distribute the 3.
\(\,\,\,\,\,6y+3-y=13\)
Combine like terms.
\(\,\,\,\,\,5y+3=13\)
Subtract 3 from both sides.
\(\,\,\,\,\,5y=10\)
Divide both sides by 5.
\(\,\,\,\,\,y=2\)
Substitute \(y=2\) into \(x=2y+1\).
\(\,\,\,\,\,x=2(2)+1\)
\(\,\,\,\,\,x=5\)
Therefore, the solution is \((5,2)\).
\(\textbf{11)}\) \(5x-2y=4\)
\(x=y+2\)
The answer is \((0,-2)\).
Substitute \(x=y+2\) into the first equation.
\(\,\,\,\,\,5(y+2)-2y=4\)
Distribute the 5.
\(\,\,\,\,\,5y+10-2y=4\)
Combine like terms.
\(\,\,\,\,\,3y+10=4\)
Subtract 10 from both sides.
\(\,\,\,\,\,3y=-6\)
Divide both sides by 3.
\(\,\,\,\,\,y=-2\)
Substitute \(y=-2\) into \(x=y+2\).
\(\,\,\,\,\,x=-2+2\)
\(\,\,\,\,\,x=0\)
Therefore, the solution is \((0,-2)\).
\(\textbf{12)}\) \(y=6-2x\)
\(4x+3y=10\)
The answer is \((4,-2)\).
Substitute \(y=6-2x\) into the second equation.
\(\,\,\,\,\,4x+3(6-2x)=10\)
Distribute the 3.
\(\,\,\,\,\,4x+18-6x=10\)
Combine like terms.
\(\,\,\,\,\,-2x+18=10\)
Subtract 18 from both sides.
\(\,\,\,\,\,-2x=-8\)
Divide both sides by \(-2\).
\(\,\,\,\,\,x=4\)
Substitute \(x=4\) into \(y=6-2x\).
\(\,\,\,\,\,y=6-2(4)\)
\(\,\,\,\,\,y=-2\)
Therefore, the solution is \((4,-2)\).
\(\textbf{13)}\) \(2x+y=-1\)
\(y=-5x+8\)
The answer is \((3,-7)\).
Substitute \(y=-5x+8\) into the first equation.
\(\,\,\,\,\,2x+(-5x+8)=-1\)
Combine like terms.
\(\,\,\,\,\,-3x+8=-1\)
Subtract 8 from both sides.
\(\,\,\,\,\,-3x=-9\)
Divide both sides by \(-3\).
\(\,\,\,\,\,x=3\)
Substitute \(x=3\) into \(y=-5x+8\).
\(\,\,\,\,\,y=-5(3)+8\)
\(\,\,\,\,\,y=-15+8\)
\(\,\,\,\,\,y=-7\)
Therefore, the solution is \((3,-7)\).
\(\textbf{14)}\) \(x=\frac{1}{2}y+3\)
\(x+y=9\)
The answer is \((5,4)\).
Substitute \(x=\frac{1}{2}y+3\) into the second equation.
\(\,\,\,\,\,\left(\frac{1}{2}y+3\right)+y=9\)
Combine like terms.
\(\,\,\,\,\,\frac{3}{2}y+3=9\)
Subtract 3 from both sides.
\(\,\,\,\,\,\frac{3}{2}y=6\)
Multiply both sides by \(\frac{2}{3}\).
\(\,\,\,\,\,y=4\)
Substitute \(y=4\) into \(x=\frac{1}{2}y+3\).
\(\,\,\,\,\,x=\frac{1}{2}(4)+3\)
\(\,\,\,\,\,x=2+3\)
\(\,\,\,\,\,x=5\)
Therefore, the solution is \((5,4)\).
\(\textbf{15)}\) \(y=0.5x+2\)
\(y=-x+8\)
The answer is \((4,4)\).
Set the two expressions for \(y\) equal to each other.
\(\,\,\,\,\,0.5x+2=-x+8\)
Add \(x\) to both sides.
\(\,\,\,\,\,1.5x+2=8\)
Subtract 2 from both sides.
\(\,\,\,\,\,1.5x=6\)
Divide both sides by \(1.5\).
\(\,\,\,\,\,x=4\)
Substitute \(x=4\) into \(y=-x+8\).
\(\,\,\,\,\,y=-4+8\)
\(\,\,\,\,\,y=4\)
Therefore, the solution is \((4,4)\).
\(\textbf{16)}\) \(y=2x-3\)
\(4x-2y=6\)
The answer is Infinitely Many Solutions or Consistent and Dependent.
Substitute \(y=2x-3\) into the second equation.
\(\,\,\,\,\,4x-2(2x-3)=6\)
Distribute the \(-2\).
\(\,\,\,\,\,4x-4x+6=6\)
Combine like terms.
\(\,\,\,\,\,6=6\)
This statement is always true.
The equations represent the same line.
Therefore, the system has infinitely many solutions.
\(\textbf{17)}\) \(y=-x+4\)
\(2x+2y=14\)
The answer is No Solution or Inconsistent.
Substitute \(y=-x+4\) into the second equation.
\(\,\,\,\,\,2x+2(-x+4)=14\)
Distribute the 2.
\(\,\,\,\,\,2x-2x+8=14\)
Combine like terms.
\(\,\,\,\,\,8=14\)
This statement is false, so the lines cannot intersect.
Therefore, the system has no solution.
\(\textbf{18)}\) The sum of two numbers is 26. One number is 4 greater than the other. Find the two numbers.
The numbers are \(11\) and \(15\).
Let the smaller number be \(x\) and the larger number be \(y\).
Write an equation for their sum.
\(\,\,\,\,\,x+y=26\)
Write an equation for the larger number.
\(\,\,\,\,\,y=x+4\)
Substitute \(y=x+4\) into the first equation.
\(\,\,\,\,\,x+(x+4)=26\)
Combine like terms.
\(\,\,\,\,\,2x+4=26\)
Subtract 4 from both sides.
\(\,\,\,\,\,2x=22\)
Divide both sides by 2.
\(\,\,\,\,\,x=11\)
Substitute \(x=11\) into \(y=x+4\).
\(\,\,\,\,\,y=11+4\)
\(\,\,\,\,\,y=15\)
Therefore, the two numbers are \(11\) and \(15\).
\(\textbf{19)}\) Adult tickets to a school play cost $8 and student tickets cost $5. A total of 40 tickets were sold for $260. How many adult tickets and student tickets were sold?
There were \(20\) adult tickets and \(20\) student tickets sold.
Let \(a\) represent the number of adult tickets and \(s\) represent the number of student tickets.
Write an equation for the total number of tickets.
\(\,\,\,\,\,a+s=40\)
Write an equation for the total amount of money.
\(\,\,\,\,\,8a+5s=260\)
Solve the first equation for \(s\).
\(\,\,\,\,\,s=40-a\)
Substitute \(s=40-a\) into the money equation.
\(\,\,\,\,\,8a+5(40-a)=260\)
Distribute the 5.
\(\,\,\,\,\,8a+200-5a=260\)
Combine like terms.
\(\,\,\,\,\,3a+200=260\)
Subtract 200 from both sides.
\(\,\,\,\,\,3a=60\)
Divide both sides by 3.
\(\,\,\,\,\,a=20\)
Substitute \(a=20\) into \(s=40-a\).
\(\,\,\,\,\,s=40-20\)
\(\,\,\,\,\,s=20\)
Therefore, 20 adult tickets and 20 student tickets were sold.
\(\textbf{20)}\) The perimeter of a rectangle is 54 feet. The length is 3 feet more than twice the width. Find the dimensions of the rectangle.
The width is \(8\) feet and the length is \(19\) feet.
Let \(L\) represent the length and \(W\) represent the width.
Use the perimeter formula.
\(\,\,\,\,\,2L+2W=54\)
The length is 3 feet more than twice the width.
\(\,\,\,\,\,L=2W+3\)
Substitute \(L=2W+3\) into the perimeter equation.
\(\,\,\,\,\,2(2W+3)+2W=54\)
Distribute the 2.
\(\,\,\,\,\,4W+6+2W=54\)
Combine like terms.
\(\,\,\,\,\,6W+6=54\)
Subtract 6 from both sides.
\(\,\,\,\,\,6W=48\)
Divide both sides by 6.
\(\,\,\,\,\,W=8\)
Substitute \(W=8\) into \(L=2W+3\).
\(\,\,\,\,\,L=2(8)+3\)
\(\,\,\,\,\,L=19\)
Therefore, the width is 8 feet and the length is 19 feet.
See Related Pages\(\)
