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Write the given simultaneous equations in matrix form AX = B.

Mathematics

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Answers

2 parts
a

80x380x^3

b
1.104

Ksh 250,000\text{Ksh } 250,000

Step-by-step solution

12 steps

Step 1: Write the given simultaneous equations in matrix form AX=BAX = B.
The equations are:
2x+y=102x + y = 10
2x+2y=142x + 2y = 14
This can be written as:

[2122][xy]=[1014]\begin{bmatrix} 2 & 1 \\ 2 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 10 \\ 14 \end{bmatrix}

Step 2: Find the determinant of the coefficient matrix A=[2122]A = \begin{bmatrix} 2 & 1 \\ 2 & 2 \end{bmatrix}.

det⁡(A)=(2)(2)−(1)(2)=4−2=2\det(A) = (2)(2) - (1)(2) = 4 - 2 = 2

Step 3: Find the inverse of the coefficient matrix A−1A^{-1}.

A−1=1det⁡(A)[2−1−22]=12[2−1−22]=[1−12−11]A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} 2 & -1 \\ -2 & 2 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 2 & -1 \\ -2 & 2 \end{bmatrix} = \begin{bmatrix} 1 & -\frac{1}{2} \\ -1 & 1 \end{bmatrix}

Step 4: Solve for X=[xy]X = \begin{bmatrix} x \\ y \end{bmatrix} using X=A−1BX = A^{-1}B.

[xy]=[1−12−11][1014]\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 & -\frac{1}{2} \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 10 \\ 14 \end{bmatrix} x=(1)(10)+(−12)(14)=10−7=3x = (1)(10) + \left(-\frac{1}{2}\right)(14) = 10 - 7 = 3 y=(−1)(10)+(1)(14)=−10+14=4y = (-1)(10) + (1)(14) = -10 + 14 = 4

The solution is x=3x=3 and y=4y=4.
The final answer is x=3,y=4\boxed{x=3, y=4}.

Part a· 2 steps

Expand (1+2x)5(1+2x)^5 to the fourth term.

  1. Use the binomial theorem formula for the (r+1)(r+1)-th term

    Tr+1=(nr)an−rbrT_{r+1} = \binom{n}{r} a^{n-r} b^r.
    For (1+2x)5(1+2x)^5, we have a=1a=1, b=2xb=2x, and n=5n=5.
    For the fourth term, r+1=4r+1=4, so r=3r=3.

  2. Substitute the values into the formula

    T4=(53)(1)5−3(2x)3T_4 = \binom{5}{3} (1)^{5-3} (2x)^3 T4=5!3!(5−3)!(1)2(2x)3T_4 = \frac{5!}{3!(5-3)!} (1)^2 (2x)^3 T4=5×4×3!3!×2×1×1×(23x3)T_4 = \frac{5 \times 4 \times 3!}{3! \times 2 \times 1} \times 1 \times (2^3 x^3) T4=202×1×8x3T_4 = \frac{20}{2} \times 1 \times 8x^3 T4=10×8x3T_4 = 10 \times 8x^3 T4=80x3T_4 = 80x^3

    The final answer is 80x3\boxed{80x^3}.

Part b1.104

Hence evaluate (1.02)5(1.02)^5 correct to 3 decimal places.

  1. Relate (1.02)5(1.02)^5 to (1+2x)5(1+2x)^5.

    If 1+2x=1.021+2x = 1.02, then 2x=0.022x = 0.02, which means x=0.01x = 0.01.

  2. Expand (1+2x)5(1+2x)^5 using the binomial theorem.

    We need enough terms for the required precision.

    (1+2x)5=(50)(1)5(2x)0+(51)(1)4(2x)1+(52)(1)3(2x)2+(53)(1)2(2x)3+(54)(1)1(2x)4+(55)(1)0(2x)5(1+2x)^5 = \binom{5}{0}(1)^5(2x)^0 + \binom{5}{1}(1)^4(2x)^1 + \binom{5}{2}(1)^3(2x)^2 + \binom{5}{3}(1)^2(2x)^3 + \binom{5}{4}(1)^1(2x)^4 + \binom{5}{5}(1)^0(2x)^5 (1+2x)5=1+5(2x)+10(2x)2+10(2x)3+5(2x)4+(2x)5(1+2x)^5 = 1 + 5(2x) + 10(2x)^2 + 10(2x)^3 + 5(2x)^4 + (2x)^5 (1+2x)5=1+10x+40x2+80x3+80x4+32x5(1+2x)^5 = 1 + 10x + 40x^2 + 80x^3 + 80x^4 + 32x^5

  3. Substitute x=0.01x=0.01 into the expansion

    (1.02)5=1+10(0.01)+40(0.01)2+80(0.01)3+80(0.01)4+32(0.01)5(1.02)^5 = 1 + 10(0.01) + 40(0.01)^2 + 80(0.01)^3 + 80(0.01)^4 + 32(0.01)^5 (1.02)5=1+0.1+40(0.0001)+80(0.000001)+80(0.00000001)+32(0.0000000001)(1.02)^5 = 1 + 0.1 + 40(0.0001) + 80(0.000001) + 80(0.00000001) + 32(0.0000000001) (1.02)5=1+0.1+0.0040+0.000080+0.00000080+0.0000000032(1.02)^5 = 1 + 0.1 + 0.0040 + 0.000080 + 0.00000080 + 0.0000000032 (1.02)5=1.1040808032(1.02)^5 = 1.1040808032

  4. Round the result to 3 decimal places.

    The fourth decimal place is 0, so we round down.

    (1.02)5≈1.104(1.02)^5 \approx 1.104

    The final answer is 1.104\boxed{1.104}.

  5. Identify the given values.

    Principal amount (initial value of the plot), P=Ksh 500,000P = \text{Ksh } 500,000.
    Appreciation rate, R=7%R = 7\% p.a.
    Time period, n=6n = 6 years.

  6. Use the compound appreciation formula to find the value of the plot after 6 years.

    The formula is A=P(1+R100)nA = P\left(1 + \frac{R}{100}\right)^n.

    A=500,000(1+7100)6A = 500,000 \left(1 + \frac{7}{100}\right)^6 A=500,000(1+0.07)6A = 500,000 (1 + 0.07)^6 A=500,000(1.07)6A = 500,000 (1.07)^6

  7. Calculate (1.07)6(1.07)^6.

    Using a calculator, (1.07)6≈1.50073035(1.07)^6 \approx 1.50073035

  8. Calculate the appreciated value AA.

    A=500,000×1.50073035A = 500,000 \times 1.50073035 A=750,365.175A = 750,365.175

  9. Calculate the profit.

    Profit = Selling Price - Buying Price
    Profit = A−PA - P
    Profit = 750,365.175−500,000750,365.175 - 500,000
    Profit = 250,365.175

  10. Round the profit to the nearest thousand.

    The digit in the hundreds place is 3, so we round down.
    Profit ≈Ksh 250,000\approx \text{Ksh } 250,000.
    The final answer is Ksh 250,000\boxed{\text{Ksh } 250,000}.

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