Step 1: Write the given simultaneous equations in matrix form $AX = B$. The equations are: $2x + y = 10$ $2x + 2y = 14$ This can be written as: $$ \begin{bmatrix} 2 & 1 \\ 2 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 10 \\ 14 \end{bmatrix} $$ Step 2: Find the determinant of the coefficient matrix $A = \begin{bmatrix} 2 & 1 \\ 2 & 2 \end{bmatrix}$. $$ \det(A) = (2)(2) - (1)(2) = 4 - 2 = 2 $$ Step 3: Find the inverse of the coefficient matrix $A^{-1}$. $$ A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} 2 & -1 \\ -2 & 2 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 2 & -1 \\ -2 & 2 \end{bmatrix} = \begin{bmatrix} 1 & -\frac{1}{2} \\ -1 & 1 \end{bmatrix} $$ Step 4: Solve for $X = \begin{bmatrix} x \\ y \end{bmatrix}$ using $X = A^{-1}B$. $$ \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 & -\frac{1}{2} \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 10 \\ 14 \end{bmatrix} $$ $$ x = (1)(10) + \left(-\frac{1}{2}\right)(14) = 10 - 7 = 3 $$ $$ y = (-1)(10) + (1)(14) = -10 + 14 = 4 $$ The solution is $x=3$ and $y=4$. The final answer is $\boxed{x=3, y=4}$. --- a) Expand $(1+2x)^5$ to the fourth term. Step 1: Use the binomial theorem formula for the $(r+1)$-th term: $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. For $(1+2x)^5$, we have $a=1$, $b=2x$, and $n=5$. For the fourth term, $r+1=4$, so $r=3$. Step 2: Substitute the values into the formula: $$ T_4 = \binom{5}{3} (1)^{5-3} (2x)^3 $$ $$ T_4 = \frac{5!}{3!(5-3)!} (1)^2 (2x)^3 $$ $$ T_4 = \frac{5 \times 4 \times 3!}{3! \times 2 \times 1} \times 1 \times (2^3 x^3) $$ $$ T_4 = \frac{20}{2} \times 1 \times 8x^3 $$ $$ T_4 = 10 \times 8x^3 $$ $$ T_4 = 80x^3 $$ The final answer is $\boxed{80x^3}$. b) Hence evaluate $(1.02)^5$ correct to 3 decimal places. Step 1: Relate $(1.02)^5$ to $(1+2x)^5$. If $1+2x = 1.02$, then $2x = 0.02$, which means $x = 0.01$. Step 2: Expand $(1+2x)^5$ using the binomial theorem. We need enough terms for the required precision. $$ (1+2x)^5 = \binom{5}{0}(1)^5(2x)^0 + \binom{5}{1}(1)^4(2x)^1 + \binom{5}{2}(1)^3(2x)^2 + \binom{5}{3}(1)^2(2x)^3 + \binom{5}{4}(1)^1(2x)^4 + \binom{5}{5}(1)^0(2x)^5 $$ $$ (1+2x)^5 = 1 + 5(2x) + 10(2x)^2 + 10(2x)^3 + 5(2x)^4 + (2x)^5 $$ $$ (1+2x)^5 = 1 + 10x + 40x^2 + 80x^3 + 80x^4 + 32x^5 $$ Step 3: Substitute $x=0.01$ into the expansion: $$ (1.02)^5 = 1 + 10(0.01) + 40(0.01)^2 + 80(0.01)^3 + 80(0.01)^4 + 32(0.01)^5 $$ $$ (1.02)^5 = 1 + 0.1 + 40(0.0001) + 80(0.000001) + 80(0.00000001) + 32(0.0000000001) $$ $$ (1.02)^5 = 1 + 0.1 + 0.0040 + 0.000080 + 0.00000080 + 0.0000000032 $$ $$ (1.02)^5 = 1.1040808032 $$ Step 4: Round the result to 3 decimal places. The fourth decimal place is $0$, so we round down. $$ (1.02)^5 \approx 1.104 $$ The final answer is $\boxed{1.104}$. --- Step 1: Identify the given values. Principal amount (initial value of the plot), $P = \text{Ksh } 500,000$. Appreciation rate, $R = 7\%$ p.a. Time period, $n = 6$ years. Step 2: Use the compound appreciation formula to find the value of the plot after 6 years. The formula is $A = P\left(1 + \frac{R}{100}\right)^n$. $$ A = 500,000 \left(1 + \frac{7}{100}\right)^6 $$ $$ A = 500,000 (1 + 0.07)^6 $$ $$ A = 500,000 (1.07)^6 $$ Step 3: Calculate $(1.07)^6$. Using a calculator, $(1.07)^6 \approx 1.50073035$ Step 4: Calculate the appreciated value $A$. $$ A = 500,000 \times 1.50073035 $$ $$ A = 750,365.175 $$ Step 5: Calculate the profit. Profit = Selling Price - Buying Price Profit = $A - P$ Profit = $750,365.175 - 500,000$ Profit = $250,365.175$ Step 6: Round the profit to the nearest thousand. The digit in the hundreds place is $3$, so we round down. Profit $\approx \text{Ksh } 250,000$. The final answer is $\boxed{\text{Ksh } 250,000}$. 3 done, 2 left today. You're making progress.