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Question 1168173: Lanre income from interest and wages is 4000 he double his investment and also get an increase to 6500 my what was his original income separately in both term (interest and wages). Using crammer rules.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let $i$ be Lanre's original income from interest, and $w$ be his original income from wages.
From the first statement, his total original income is $4000:
Equation 1: $i + w = 4000$
From the second statement, he doubles his investment (which we assume directly doubles his interest income) and his total income becomes $6500. His wage income remains the same.
New interest income = $2i$
New wage income = $w$
New total income = $6500$
Equation 2: $2i + w = 6500$
We have a system of two linear equations with two variables:
1) $i + w = 4000$
2) $2i + w = 6500$
We will use Cramer's rule to solve this system. Cramer's rule states that for a system of linear equations:
$a_1x + b_1y = c_1$
$a_2x + b_2y = c_2$
The solutions for $x$ and $y$ are given by:
$x = \frac{D_x}{D}$
$y = \frac{D_y}{D}$
where $D$ is the determinant of the coefficient matrix, and $D_x$ and $D_y$ are the determinants of matrices formed by replacing the column corresponding to $x$ and $y$ respectively with the constant terms.
In our case, the variables are $i$ and $w$:
$1i + 1w = 4000$
$2i + 1w = 6500$
Here, $a_1 = 1, b_1 = 1, c_1 = 4000$ and $a_2 = 2, b_2 = 1, c_2 = 6500$.
First, calculate the determinant of the coefficient matrix $D$:
$D = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} = (1 \times 1) - (1 \times 2) = 1 - 2 = -1$
Next, calculate the determinant $D_i$ by replacing the first column (coefficients of $i$) with the constant terms:
$D_i = \begin{vmatrix} 4000 & 1 \\ 6500 & 1 \end{vmatrix} = (4000 \times 1) - (1 \times 6500) = 4000 - 6500 = -2500$
Now, calculate the determinant $D_w$ by replacing the second column (coefficients of $w$) with the constant terms:
$D_w = \begin{vmatrix} 1 & 4000 \\ 2 & 6500 \end{vmatrix} = (1 \times 6500) - (4000 \times 2) = 6500 - 8000 = -1500$
Finally, use Cramer's rule to find the values of $i$ and $w$:
$i = \frac{D_i}{D} = \frac{-2500}{-1} = 2500$
$w = \frac{D_w}{D} = \frac{-1500}{-1} = 1500$
So, Lanre's original income from interest was $2500, and his original income from wages was $1500.
Let's check if these values satisfy the given conditions:
Original total income: $2500 + 1500 = 4000$ (Correct)
Doubled investment (interest) and increased total income: $2(2500) + 1500 = 5000 + 1500 = 6500$ (Correct)
Final Answer: The final answer is $\boxed{Original interest income was $2500 and original wage income was $1500}$
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