Quick help: Explain how to solve the following system of equations. What is the solution to the system?

2 answers:

7 0

-3x - 4y - 2z = 0
x + 3y + 2z = 1

-2x - y = 1

-4x - 4y - 2z = 10
3x + 4y +2z= 0

-x = 10. x = -10

-2(-10) - y = 1
20 - y = 1
-y = -19
y = 19

-10 + 3(19) + 2z = 1
-10 + 57 + 2z = 1
47 + 2z = 1
2z = -46
z= -23

check: 2(-10)+2(19)-23=-5
-20+38-23=-5
-43+38=-5
-5=-5

lana66690 [7]4 years ago

5 0

Given equations:
2x+2y+z = -5 -----------(1)
3x+4y+2z = 0 ----------(2)
x+3y+2z = 1 ---------(3)
Subtract equation (3) from equation (2) to eliminate z
3x+4y+2z =0
-x-3y-2z=-1
__________
2x+y=-1 -------------------(4)
Multiply equation (1) by 2 and subtract it from equation (3) to eliminate z
x+3y+2z=1
-4x-4y-2z=10 ____________
-3x-y=11 --------(5)
Add equation (4) and (5) to eliminate y
2x+y=-1
-3x-y=11 _______
-x=10
X=-10
Substitute the value of x in equation (4) to find y
2(-10)+y=-1
-20+y=-1
y=-1+20
y=19
Substitute the values of x and y in any one of the 3 equations, to find z.
Letâ€™s substitute in equation (1)
2(-10)+2(19)+z=-5
-20+38+z=-5
18+z=-5
z=-5-18
z=-23
Therefore the solution is: x=-10, y=19, z=-23

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The function f(x)=(x-5)^2 +2 is not one-to-one. Identify a restricted domain that makes the function one-to-one, and find the in

Anon25 [30]

We have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

We know that vertex of this quadratic function occurs at (5,2).

Further, we know that range of this function is $[2,\infty)$ .

If we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

Let us know find its inverse.

$y=(x-5)^{2}+2$

Upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

Let us now solve this function for y.

$(y-5)^{2}=x-2\\
y-5=\pm \sqrt{x-2}\\
y=5\pm \sqrt{x-2}\\$

Hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .