Help Patulong linear algebra

[Rishi AFK]

 

[Chiiiin]

1. Solve each given linear system

Problem 1 has three parts, each with a different system of linear equations. Let's solve each system step by step.

For part (a):
[ \begin{align*} x + y &= 3 \quad \text{(i)} \ x - y &= -3 \quad \text{(ii)} \ \end{align*} ]

We can add equation (i) and (ii) to eliminate ( y ) and solve for ( x ). Then we can substitute the value of ( x ) into either equation to find ( y ).

For part (b):
[ \begin{align*} x - 2y + 3z &= 9 \quad \text{(iii)} \ x - 2y + z &= -5 \quad \text{(iv)} \ 3x + y + 2z &= 1 \quad \text{(v)} \ \end{align*} ]

We will need to use either substitution or elimination to solve this system. It's a 3x3 system, so it might involve a few more steps.

For part (c):
[ \begin{align*} \frac{x}{6} + \frac{y}{6} &= 1 \quad \text{(vi)} \ x - y &= -3 \quad \text{(vii)} \ \end{align*} ]

Equation (vi) can be simplified by multiplying through by 6 to eliminate the fractions, and then we can proceed with solving the system similarly to part (a).

For part (a) of problem 1, the solution to the system of equations is ( x = 0 ) and ( y = 3 ).
Let's move on to part (b). We will solve the 3x3 system using either substitution or elimination. I will calculate this next.

For part (b) of problem 1, the solution to the system of equations is ( x = -\frac{38}{7} ), ( y = \frac{23}{7} ), and ( z = 7 ).
Now, let's handle part (c). I'll perform the calculation for this final part.

For part (c) of problem 1, the solution to the system of equations is ( x = \frac{3}{2} ) and ( y = \frac{9}{2} ).

 

[Jules_14]

use caltech easy lang yan

 

[Wu_Fei]

YøùTùbé mas madali.. saka sipag need jan.. di agad matutunan yan.. tyaga sipag puhunan jan.. kahit saan naman sipag tyaga puhunan

 

[Kene Seu]

try ch@t6pT ts