a) Using Cramer's rule, solve the following equations: 2x + 5y = 17 , 5x 2y = 1 . [2] b) Find the real numbers x and y if (x 1)i + (y + 1) = (1 + i)(4 3i) . [2] c) Find the value of K so that the…

a) Solving the system of equations using Cramer’s Rule

Given the system of equations: $$2x+5y=17\text{(1)}$$ $$5x-2y=-1\text{(2)}$$

Step 1: Write the system in matrix form $AX=B$

The coefficient matrix $A$, variable matrix $X$, and constant matrix $B$ are: $$A=\left[\begin{array}{cc}2& 5\\ 5& -2\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right],B=\left[\begin{array}{c}17\\ -1\end{array}\right]$$

Step 2: Compute the determinant of $A$ ($\Delta$)

$$\Delta =\left|\begin{array}{cc}2& 5\\ 5& -2\end{array}\right|=(2)(-2)-(5)(5)=-4-25=-29$$

Step 3: Compute ${\Delta }_x$ (replace the first column of $A$ with $B$)

$${\Delta }_x=\left|\begin{array}{cc}17& 5\\ -1& -2\end{array}\right|=(17)(-2)-(5)(-1)=-34+5=-29$$

Step 4: Compute ${\Delta }_y$ (replace the second column of $A$ with $B$)

$${\Delta }_y=\left|\begin{array}{cc}2& 17\\ 5& -1\end{array}\right|=(2)(-1)-(17)(5)=-2-85=-87$$

Step 5: Solve for $x$ and $y$

$$x=\frac{{\Delta }_x}{\Delta }=\frac{-29}{-29}=1$$ $$y=\frac{{\Delta }_y}{\Delta }=\frac{-87}{-29}=3$$

Final Answer: $$\boxed{x=1,y=3}$$

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b) Finding real numbers $x$ and $y$ from the complex equation

Given: $$(x-1)i+(y+1)=(1+i)(4-3i)$$

Step 1: Expand the right-hand side (RHS)

$$(1+i)(4-3i)=1\cdot 4+1\cdot (-3i)+i\cdot 4+i\cdot (-3i)$$ $$=4-3i+4i-3i^2$$ Since $i^2=-1$: $$=4-3i+4i+3=(4+3)+(-3i+4i)=7+i$$

Step 2: Equate real and imaginary parts

The left-hand side (LHS) is: $$(x-1)i+(y+1)=(y+1)+(x-1)i$$ Equate real and imaginary parts with RHS: $$\text{Real part: }y+1=7$$ $$\text{Imaginary part: }x-1=1$$

Step 3: Solve for $x$ and $y$

$$y+1=7⟹y=6$$ $$x-1=1⟹x=2$$

Final Answer: $$\boxed{x=2,y=6}$$

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c) Finding $K$ for one root equal to zero

Given the quadratic equation: $$3x^2+7x+6-K=0$$

Step 1: Condition for one root to be zero

If $x=0$ is a root, substitute $x=0$ into the equation: $$3(0{)}^2+7(0)+6-K=0$$ $$6-K=0⟹K=6$$

Verification:

Substitute $K=6$ back into the equation: $$3x^2+7x+0=0$$ $$x(3x+7)=0$$ This gives roots $x=0$ and $x=-\frac{7}{3}$, confirming one root is zero.

Final Answer: $$\boxed{K=6}$$