126=M(.03)+(4000-M)*.04)
where M is at 3 percent.
126=m(.03-.04)+160
M= 34/.01= 3400
and 4000-M=600
where M is at 3 percent.
126=m(.03-.04)+160
M= 34/.01= 3400
and 4000-M=600
Let's denote the amount invested in the account earning 3% interest as "x" and the amount invested in the account earning 4% interest as "y".
According to the problem, we know that the total amount invested is $4,000, so we can express this information as:
x + y = 4000 -- Equation 1
We also know that the total interest earned for the year is $126. Since interest is calculated by multiplying the principal by the interest rate, we can calculate the interest earned in each account as:
0.03x + 0.04y = 126 -- Equation 2
Now, we can solve this system of equations to find the values of x and y.
One way to do this is by using the substitution method:
1. Solve Equation 1 for y:
y = 4000 - x
2. Substitute the value of y in Equation 2:
0.03x + 0.04(4000 - x) = 126
3. Simplify and solve for x:
0.03x + 160 - 0.04x = 126
-0.01x = -34
x = -34 / -0.01
x = 3400
Now that we have found the value of x, we can substitute it back into Equation 1 to find the value of y:
x + y = 4000
3400 + y = 4000
y = 4000 - 3400
y = 600
Therefore, $3,400 is invested in the account earning 3% interest, and $600 is invested in the account earning 4% interest.