What is the solution to the system? 1. 5 x -4 y -3 z =3, 2. x + y -z =0, 3. x = 3 y + 1? i haven't done this kind of math. i am trying to my really doesn't know how to do it. we are each other.

(1) 5 x -4 y -3 z =3

(2) x + y -z =0

(3) x = 3 y + 1

We can use substitution, when we take the equation x = 3 y + 1, and substitute x in the equations (1) and (2) by 3y+1.

(1) 5 x -4 y -3 z =3

5(3y+1) -4 y -3 z =3 ---> 15y+5-4y-3z=3 > 11y - 3z = - 2 (4)

(2) x + y -z =0 > (2) (3y + 1) + y -z =0 > 4y - z+1 > z= 4y+1 (5)

Now, we can take the equation (5) and substitute the value of (z) into the equation (4).

(4) 11y - 3z = - 2

(5) z= 4y+1

(6) 11y - 3z = - 2 ---> 11y - 3(4y+1) = - 2 --->11y -12y -3 = -2 ---> -y= 1> y= - 1

Take the equation (5) , and now we know y = - 1.

(5) z= 4y+1 =4*(-1)+1 = -4 +1= - 3, > z= - 3

Take the equation (3), and we know that y = - 1,

x = 3 y + 1 = 3*(-1) + 1 = -3 +1 = -2, x = -2.

So, x = - 2, y = - 1, z = - 3.

Check:

(1) 5 *(-2) -4*(-1) -3*(-3) =3 ---> -10+4+9=3---> 3=3

(2) (-2) + (-1) -(-3) =0 ---> -2-1+3=0 > 0=0

(3) (-2) = 3*(-1) + 1 > -2=-3+1 ---> -2 = -2

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

27 +27 or 27times 2=54

the diffrence is 0