What Is The Student ID Of Nur Uddin Vai?

$-3x+2y-6z=6$ $\color{#3D99F6}(1)$

$5x+7y-5z=6$ $\color{#3D99F6}(2)$

$x+4y-2z=8$ $\color{#3D99F6}(3)$

Multiply $\color{#3D99F6}(1)$ by $5$ and multiply $\color{#3D99F6}(2)$ by $3$ then add the results: $\implies$ $31y-45z=48$ $\color{#3D99F6}(4)$

Multiply $\color{#3D99F6}(3)$ by $-5$ then add the result to $\color{#3D99F6}(2)$ : $\implies$ $-13y+5z=-34$ $\color{#3D99F6}(5)$

Multiply $\color{#3D99F6}(4)$ by $13$ and multiply $\color{#3D99F6}(5)$ by $31$ then add the results: $\implies$ $-430z=430$ $\implies$ $\boxed{z=1}$

x+y=z

1. Substitute (x+y) for z into two of the equations

-3x+2y- 6z=6

-3x+2y- 6(x+y)=6

-3x+2y-6x-6y=6

-9x-4y=6

x+4y-2z=8

x+4y-2(x+y)=8

x+4y-2x-2y=8

-x+2y=8

2. Next, we use elimination.

-9x-4y=6

-x+2y=8

Multiply the second equation by 2 to get the y terms to cancel.

2(-x-2y)=2(8)

-2x+4y=16

-9x-4y=6

-2x+4y=16

Y terms cancel

-9x=6

-2x=16

-9x-2x=6+16

-11x=22

x = -2

3. Substitute -2 for x to solve for y

-x+2y=8

-(-2)+2y=8

2+2y=8

2y=6

y=3

x+y=z

-2+3=z

1=z

Multiply second equation by 3 and third by 2 you will get 15x + 21y - 15z = 18
2x + 8y - 4z = -16
Subtract second from first. You will get 13x + 13y - 11z = 2
13 (x + y) - 11z = 2
13z - 11z = 2, 2z = 2, z = 1

The real trick here was to manipulate the two equations in such a way that x and y coefficients are equal and that again is no rocket science but pure maths. If we look at the coefficients we see that in first equation y is 2 more than x and in second it is 3 more. We cannot get them equal by adding them so we have to subtract one from other. Now comes the intelligent part. So if we multiply first equation with 3 then y would be 6 more than x. And when we reverse the signs and multiply the second one by 2, x will be 6 more than y. Now if we add together the coefficients will be equal.

$\begin{cases} -3x + 2y -6z = 6 \\ 5x + 7y - 5z = 6 \\ x + 4y - 2z = 8 \end{cases}$ Solving: $-15x + 10y - 30z = 30$ $15x + 21y - 15z = 18$ $31y - 45z = 48$ $-5x - 20y + 10z = -40$ $-13y + 5z = -34$ $403y - 585z = 624$ $403y + 155z = 1054$ $-430z = 430$ $z = -1$

$-3x+2y-6z=6$ $\color{#20A900}(1)$

$5x+7y-5z=6$ $\color{#20A900}(2)$

$x+4y-2z=8$ $\color{#20A900}(3)$

Multiply $\color{#20A900}(1)$ by $5$ and $\color{#20A900}(2)$ by $3$ , then add the results to eliminate $x$

$5(-3x+2y-6z=6)$ $\implies$ $-15x+10y-30z=30$ $\color{#20A900}(4)$

$3(5x+7y-5z=6)$ $\implies$ $15x+21y-15z=18$ $\color{#20A900}(5)$

Add the results

$\color{#20A900}(4)$ $+$ $\color{#20A900}(5)$ $\implies$ $31y-45z=48$ $\color{#20A900}(6)$

Now, we must eliminate $x$ in $\color{#20A900}(2)$ and $\color{#20A900}(3)$ . Multiply $\color{#20A900}(3)$ by $-5$ then add to $\color{#20A900}(2)$ to eliminate $x$ .

$-5(x+4y-2z=8)$ $\implies$ $-5x-20y+10z=-40$ $\color{#20A900}(7)$

$\color{#20A900}(2)$ $+$ $\color{#20A900}(7)$ $\implies$ $-13y+5z=-34$ $\color{#20A900}(8)$

Multiply $\color{#20A900}(8)$ by $9$ then add to $\color{#20A900}(6)$ to eliminate $z$ and solve for $y$ .

$9(-13y+5z=-34)$ $\implies$ $-117y+45z=-306$ $\color{#20A900}(10)$

$\color{#20A900}(6)$ $+$ $\color{#20A900}(10)$ $\implies$ $-86y=-258$ $\implies$ $\boxed{\color{#D61F06}\large y=3}$

Solve for $z$ in $\color{#20A900}(8)$ by substituting $3$ for $y$ .

$-13(3)+5z=-34$ $\implies$ $-39+5z=-34$ $\implies$ $\boxed{\color{#D61F06}\large z=1}$

Solve for $x$ in $\color{#20A900}(1)$ or $\color{#20A900}(2)$ or $\color{#20A900}(3)$ . In here, I used $\color{#20A900}(3)$ because there is no coefficient of $x$

$x+4(3)-2(1)=8$ $\implies$ $\boxed{\large\color{#D61F06}\large x=-2}$