Answer :
Sure, let’s solve this problem step-by-step.
Problem Statement:
Dawn has been using two bank accounts to save money for a car. The difference between account 1 and account 2 is [tex]$100. If she uses \(\frac{3}{8}\) of account 1 and \(\frac{7}{8}\) of account 2, Dawn will have a down payment of $[/tex]2000. Find the total amount of money Dawn has in each account.
Equations to solve:
[tex]\[ A - B = 100 \][/tex]
[tex]\[ \frac{3}{8}A + \frac{7}{8}B = 2000 \][/tex]
Step-by-Step Solution:
1. Equation 1:
[tex]\[ A - B = 100 \][/tex]
2. Equation 2:
[tex]\[ \frac{3}{8}A + \frac{7}{8}B = 2000 \][/tex]
To eliminate the fractions in Equation 2, multiply the entire equation by 8:
[tex]\[ 8 \left( \frac{3}{8}A + \frac{7}{8}B \right) = 8 \cdot 2000 \][/tex]
Simplifying, we get:
[tex]\[ 3A + 7B = 16000 \][/tex]
Now, we have a new simpler system of equations:
[tex]\[ A - B = 100 \][/tex]
[tex]\[ 3A + 7B = 16000 \][/tex]
3. Solving the system using substitution or elimination:
First, solve Equation 1 for [tex]\(A\)[/tex]:
[tex]\[ A = B + 100 \][/tex]
Next, substitute [tex]\(A = B + 100\)[/tex] into the second equation:
[tex]\[ 3(B + 100) + 7B = 16000 \][/tex]
[tex]\[ 3B + 300 + 7B = 16000 \][/tex]
[tex]\[ 10B + 300 = 16000 \][/tex]
Subtract 300 from both sides:
[tex]\[ 10B = 15700 \][/tex]
Divide by 10:
[tex]\[ B = 1570 \][/tex]
Now, substitute [tex]\(B = 1570\)[/tex] back into [tex]\(A = B + 100\)[/tex]:
[tex]\[ A = 1570 + 100 \][/tex]
[tex]\[ A = 1670 \][/tex]
Therefore, Dawn has \[tex]$1670 in account 1 and \$[/tex]1570 in account 2.
Problem Statement:
Dawn has been using two bank accounts to save money for a car. The difference between account 1 and account 2 is [tex]$100. If she uses \(\frac{3}{8}\) of account 1 and \(\frac{7}{8}\) of account 2, Dawn will have a down payment of $[/tex]2000. Find the total amount of money Dawn has in each account.
Equations to solve:
[tex]\[ A - B = 100 \][/tex]
[tex]\[ \frac{3}{8}A + \frac{7}{8}B = 2000 \][/tex]
Step-by-Step Solution:
1. Equation 1:
[tex]\[ A - B = 100 \][/tex]
2. Equation 2:
[tex]\[ \frac{3}{8}A + \frac{7}{8}B = 2000 \][/tex]
To eliminate the fractions in Equation 2, multiply the entire equation by 8:
[tex]\[ 8 \left( \frac{3}{8}A + \frac{7}{8}B \right) = 8 \cdot 2000 \][/tex]
Simplifying, we get:
[tex]\[ 3A + 7B = 16000 \][/tex]
Now, we have a new simpler system of equations:
[tex]\[ A - B = 100 \][/tex]
[tex]\[ 3A + 7B = 16000 \][/tex]
3. Solving the system using substitution or elimination:
First, solve Equation 1 for [tex]\(A\)[/tex]:
[tex]\[ A = B + 100 \][/tex]
Next, substitute [tex]\(A = B + 100\)[/tex] into the second equation:
[tex]\[ 3(B + 100) + 7B = 16000 \][/tex]
[tex]\[ 3B + 300 + 7B = 16000 \][/tex]
[tex]\[ 10B + 300 = 16000 \][/tex]
Subtract 300 from both sides:
[tex]\[ 10B = 15700 \][/tex]
Divide by 10:
[tex]\[ B = 1570 \][/tex]
Now, substitute [tex]\(B = 1570\)[/tex] back into [tex]\(A = B + 100\)[/tex]:
[tex]\[ A = 1570 + 100 \][/tex]
[tex]\[ A = 1670 \][/tex]
Therefore, Dawn has \[tex]$1670 in account 1 and \$[/tex]1570 in account 2.