Solve the system of linear equations given below using substitution.
Suppose there is a piggy bank that contains 57 coins, which are only quarters and dimes. The total number of coins in the bank is 57, and the total value of these coins is $9.45. This information can be represented by the following system of equations:
D + Q = 57
00.10D + 0.25Q = 9.45
Determine how many of the coins are quarters and how many are dimes.
Answer
The information for this example is represented by the following system of equations:
D + Q = 57
0.10D + 0.25Q = 9.45
Determine how many of the coins are
quarters and how many are dimes.
D = 32 (number of dimes)
Q = 25 (number of quarters
Detailed Answer
To solve this problem using substitution we need to follow the steps outlined in this section.
1. Choose one equation and isolate one variable; this equation will be considered the first equation.
The equation D + Q = 57 is one that can be easily solved for D.
We want to isolate D, so we subtract Q from both sides of the equation.
D + Q = 57
D + Q – Q = 57 – Q
D = 57 – Q
2. Substitute the solution from step 1 into the second equation and solve for the variable in the equation.
Now we substitute the value for D, which is
57 – Q, into the other equation, 0.10D + 0.25Q = 9.45.
This leaves us with one equation with only one variable, Q. We find a numeric value for Q by isolating Q.
0.10D + 0.25Q = 9.45, and
D = 57 – Q, so
0.10 (57 – Q) + 0.25Q = 9.45
First we must get rid of the parenthesis using the distributive property. Then we combine like terms.
5.7 – 0.10Q + 0.25Q = 9.45
5.7 + 0.15Q = 9.45
Now subtract 5.7 from both sides of these equation to isolate the term containing the variable Q.
5.7 + 0.15Q – 5.7 = 9.45 –5.7
0.15Q = 3.75
Divide both sides by 0.15.
(0.15/0.15)Q = 3.75/0.15
Q = 25
3. Using the value found in step 2, substitute it into the first equation and solve for the second variable.
We found that Q = 25, so we substitute that into the equation D + Q = 57.
When we do this we find D = 32.
D + Q = 57
D + 25 = 57
D + 25 – 25 = 57 – 25
D = 32
4. Substitute the values for both variables into both equation to show they are correct.
Now we should substitute the value of D = 32 and Q = 25 into both of our original equations.
D + Q = 57
32 + 25 = 57
57 = 57
0.10D + 0.25Q = 9.45
.10(32) + .25(25) = 9.45
3.20 + 6.25 = 9.45
9.45 = 9.45
These values work in both equations so we have the correct answer.
Suppose there is a piggy bank that contains 57 coins, which are only quarters and dimes. The total number of coins in the bank is 57, and the total value of these coins is $9.45. This information can be represented by the following system of equations:
D + Q = 57
00.10D + 0.25Q = 9.45
Determine how many of the coins are quarters and how many are dimes.
Answer
The information for this example is represented by the following system of equations:
D + Q = 57
0.10D + 0.25Q = 9.45
Determine how many of the coins are
quarters and how many are dimes.
D = 32 (number of dimes)
Q = 25 (number of quarters
Detailed Answer
To solve this problem using substitution we need to follow the steps outlined in this section.
1. Choose one equation and isolate one variable; this equation will be considered the first equation.
The equation D + Q = 57 is one that can be easily solved for D.
We want to isolate D, so we subtract Q from both sides of the equation.
D + Q = 57
D + Q – Q = 57 – Q
D = 57 – Q
2. Substitute the solution from step 1 into the second equation and solve for the variable in the equation.
Now we substitute the value for D, which is
57 – Q, into the other equation, 0.10D + 0.25Q = 9.45.
This leaves us with one equation with only one variable, Q. We find a numeric value for Q by isolating Q.
0.10D + 0.25Q = 9.45, and
D = 57 – Q, so
0.10 (57 – Q) + 0.25Q = 9.45
First we must get rid of the parenthesis using the distributive property. Then we combine like terms.
5.7 – 0.10Q + 0.25Q = 9.45
5.7 + 0.15Q = 9.45
Now subtract 5.7 from both sides of these equation to isolate the term containing the variable Q.
5.7 + 0.15Q – 5.7 = 9.45 –5.7
0.15Q = 3.75
Divide both sides by 0.15.
(0.15/0.15)Q = 3.75/0.15
Q = 25
3. Using the value found in step 2, substitute it into the first equation and solve for the second variable.
We found that Q = 25, so we substitute that into the equation D + Q = 57.
When we do this we find D = 32.
D + Q = 57
D + 25 = 57
D + 25 – 25 = 57 – 25
D = 32
4. Substitute the values for both variables into both equation to show they are correct.
Now we should substitute the value of D = 32 and Q = 25 into both of our original equations.
D + Q = 57
32 + 25 = 57
57 = 57
0.10D + 0.25Q = 9.45
.10(32) + .25(25) = 9.45
3.20 + 6.25 = 9.45
9.45 = 9.45
These values work in both equations so we have the correct answer.
