An amount of $5000 is put into three investmesnts at rates of 6%, 7%, and 8% per annum, respectively. The total income is $358. The income from the first two investsments is $70 more than the third investment. Find the amount of each investment.
X at 6%
Y at 7%
Z at 8%
X + Y + Z = 5000
0.06X + 0.07Y + 0.08Z = 358
0.06X + 0.07Y - 0.08Z = 70
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Subtracting the last two equations tells you that 0.16Z = 288
Z = 1800.
X + Y = 3200
0.12X + 0.14Y = 428
Take it from there. Cheack my math
X=1000. y= 2200 z =1800
To solve this problem, we can use a system of equations. Let's represent the amount invested in the first, second, and third investments as x, y, and z, respectively.
We know that the total amount invested is $5000, so the first equation is:
x + y + z = 5000
We also know that the total income from the investments is $358, so the second equation is:
0.06x + 0.07y + 0.08z = 358
Finally, we know that the income from the first two investments is $70 more than the income from the third investment, so the third equation is:
0.06x + 0.07y = 0.08z + 70
Now, we have a system of three equations that we can solve simultaneously to find the values of x, y, and z.
There are several methods to solve this system of equations, such as substitution or elimination. Let's use substitution in this case.
First, rearrange the third equation to isolate one variable. Subtract 0.08z from both sides:
0.06x + 0.07y - 0.08z = 70
Now, solve the first equation for x:
x = 5000 - y - z
Substitute this expression for x in the third equation:
0.06(5000 - y - z) + 0.07y - 0.08z = 70
Simplify:
300 - 0.06y - 0.06z + 0.07y - 0.08z = 70
Combine like terms:
0.01y - 0.14z = -230
Now, we have a system of two equations:
0.01y - 0.14z = -230 (Equation 4)
0.06x + 0.07y + 0.08z = 358 (Equation 2)
Let's solve this system using the substitution method.
Solve Equation 4 for y:
0.01y = 0.14z - 230
Multiply by 100 to eliminate the decimal:
y = 14z - 23000 (Equation 5)
Substitute this expression for y in Equation 2:
0.06x + 0.07(14z - 23000) + 0.08z = 358
Simplify:
0.06x + 0.98z - 1610 + 0.08z = 358
Combine like terms:
0.06x + 1.06z = 1968 (Equation 6)
Now, we have a system of two equations:
0.06x + 1.06z = 1968 (Equation 6)
y = 14z - 23000 (Equation 5)
We can solve this system using any method of solving systems of linear equations. Let's use elimination.
Multiply Equation 6 by 100 to eliminate the decimal:
6x + 106z = 196800 (Equation 7)
Now, multiply Equation 5 by 6:
6y = 84z - 138000
Simplify:
6y - 84z = -138000 (Equation 8)
Multiply Equation 5 by 106:
106y = 1496z - 243800
Simplify:
106y - 1496z = -243800 (Equation 9)
Now, we have a new system of two equations:
106y - 1496z = -243800 (Equation 9)
6y - 84z = -138000 (Equation 8)
Subtract Equation 8 from Equation 9 to eliminate y:
(106y - 1496z) - (6y - 84z) = -243800 - (-138000)
Simplify:
100y - 1412z = -105800
Divide the equation by 4:
25y - 353z = -26450 (Equation 10)
Now, we have a system of two equations:
25y - 353z = -26450 (Equation 10)
6y - 84z = -138000 (Equation 8)
Multiply Equation 10 by 6:
150y - 2118z = -158700 (Equation 11)
Now, subtract Equation 8 from Equation 11 to eliminate y:
(150y - 2118z) - (6y - 84z) = -158700 - (-138000)
Simplify:
144y - 2034z = -20700
Divide the equation by 6:
24y - 339z = -3450 (Equation 12)
Now, we have a new system of two equations:
24y - 339z = -3450 (Equation 12)
25y - 353z = -26450 (Equation 10)
Multiply Equation 12 by 25:
600y - 8475z = -86250 (Equation 13)
Now, subtract Equation 10 from Equation 13 to eliminate y:
(600y - 8475z) - (25y - 353z) = -86250 - (-26450)
Simplify:
575y - 8122z = -59800
Divide the equation by 19:
25y - 134z = -2545 (Equation 14)
Now, we have a new system of two equations:
25y - 134z = -2545 (Equation 14)
25y - 353z = -26450 (Equation 10)
Subtract Equation 14 from Equation 10 to eliminate y:
(25y - 353z) - (25y - 134z) = -26450 - (-2545)
Simplify:
219z = -23805
Divide both sides of the equation by 219 to solve for z:
z = -23805 / 219
Simplify:
z ≈ -108.9
Since the amount invested cannot be negative, we discard this solution.
Therefore, there is no valid solution for the amounts invested in the three investments that satisfy the given conditions.