Question

question 1
1.1 find only y using cramérs rule given the following simultaneous equations
3x - 2y = z + 2
-3y + z = x + 10
2x + 3y = 2z - 14
(8)
1.2 solve for x if |x -2|
|7 3 - x| = -26
(4)
1.3 given r=(2a - 3b)^6. use the binomial theorem to write down:
1.3.1 the first term,
1.3.2 the middle term and
1.3.3 the last term in the series expansion of r.
(3)
1.4 use the binomial theorem to expand ∛(1 + 2x) to three terms.
(4)
19

Explanation:

Step1: Rearrange equations for 1.1

Rearrange the system of equations to the standard form $a_1x + b_1y + c_1z=d_1$, $a_2x + b_2y + c_2z=d_2$, $a_3x + b_3y + c_3z=d_3$.
$3x-2y - z=2$
$-x-3y + z = 10$
$2x+3y-2z=-14$

Step2: Calculate the determinant $D$ for 1.1

$D=

$$\begin{vmatrix} 3&-2&-1\\ -1&-3&1\\ 2&3&-2 \end{vmatrix}$$

=3\times

$$\begin{vmatrix}-3&1\\3&-2\end{vmatrix}$$

-(-2)\times

$$\begin{vmatrix}-1&1\\2&-2\end{vmatrix}$$

+(-1)\times

$$\begin{vmatrix}-1&-3\\2&3\end{vmatrix}$$

$
$=3((-3)\times(-2)-1\times3)+2((-1)\times(-2)-1\times2)-1((-1)\times3-(-3)\times2)$
$=3(6 - 3)+2(2 - 2)-1(-3 + 6)=9+0 - 3=6$

Step3: Calculate the determinant $D_y$ for 1.1

Replace the second - column of $D$ with the constants on the right - hand side of the equations.
$D_y=

$$\begin{vmatrix} 3&2&-1\\ -1&10&1\\ 2&-14&-2 \end{vmatrix}$$

=3\times

$$\begin{vmatrix}10&1\\-14&-2\end{vmatrix}$$

-2\times

$$\begin{vmatrix}-1&1\\2&-2\end{vmatrix}$$

+(-1)\times

$$\begin{vmatrix}-1&10\\2&-14\end{vmatrix}$$

$
$=3(10\times(-2)-1\times(-14))-2((-1)\times(-2)-1\times2)-1((-1)\times(-14)-10\times2)$
$=3(-20 + 14)-2(2 - 2)-1(14 - 20)=3\times(-6)-0+6=-18 + 6=-12$

Step4: Find $y$ for 1.1

By Cramer's rule, $y=\frac{D_y}{D}=\frac{-12}{6}=-2$

Step5: Solve for $x$ in 1.2

Calculate the determinant: $

$$\begin{vmatrix}x&-2\\7&3 - x\end{vmatrix}$$

=x(3 - x)-(-2)\times7$
$=3x-x^{2}+14$
Set it equal to - 26: $-x^{2}+3x + 14=-26$
$x^{2}-3x - 40 = 0$
Factor the quadratic equation: $(x - 8)(x+5)=0$
So $x = 8$ or $x=-5$

Step6: Use binomial theorem for 1.3.1

The binomial theorem is $(a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}$. For $R=(2a-3b)^6$, the first term when $k = 0$ is $\binom{6}{0}(2a)^{6}(-3b)^{0}=64a^{6}$

Step7: Use binomial theorem for 1.3.2

$n = 6$, the middle term is when $k = 3$. $\binom{6}{3}(2a)^{6 - 3}(-3b)^{3}=\frac{6!}{3!(6 - 3)!}\times8a^{3}\times(-27b^{3})=20\times8a^{3}\times(-27b^{3})=-4320a^{3}b^{3}$

Step8: Use binomial theorem for 1.3.3

The last term when $k = 6$ is $\binom{6}{6}(2a)^{0}(-3b)^{6}=729b^{6}$

Step9: Use binomial theorem for 1.4

Rewrite $\sqrt[3]{1 + 2x}=(1 + 2x)^{\frac{1}{3}}$. By the binomial theorem $(1 + y)^n=1+ny+\frac{n(n - 1)}{2!}y^{2}+\cdots$ with $y = 2x$ and $n=\frac{1}{3}$
$(1 + 2x)^{\frac{1}{3}}=1+\frac{1}{3}(2x)+\frac{\frac{1}{3}(\frac{1}{3}-1)}{2}(2x)^{2}$
$=1+\frac{2}{3}x+\frac{\frac{1}{3}\times(-\frac{2}{3})}{2}\times4x^{2}=1+\frac{2}{3}x-\frac{4}{9}x^{2}$

Answer:

1.1: $y=-2$
1.2: $x = 8$ or $x=-5$
1.3.1: $64a^{6}$
1.3.2: $-4320a^{3}b^{3}$
1.3.3: $729b^{6}$
1.4: $1+\frac{2}{3}x-\frac{4}{9}x^{2}$