Solve the system of equations.
3x + 4y + 6z = 3
4x + 3y + 3z - 4
5x + 6y + 7z = 5
a. (x = 0, y = 1,z = 1)
b. (x = 2, y = -1,z=-1)
c. (x=3, y=-2,z=2)
d. (x = 1, y = 0,z=0)

Value of x = 1, y = 0 and z = 0

Option D is correct.

Step-by-step explanation:

3x + 4y + 6z = 3

4x + 3y + 3z = 4

5x + 6y + 7z = 5

We need to solve and find values of x, y and z.

I am using Elimination method.

Let

3x + 4y + 6z = 3    (1)

4x + 3y + 3z= 4     (2)

5x + 6y + 7z = 5   (3)

Multiply eq(1) with 4 and eq(2) with 3 and subtracting

12x+16y+24z = 12

12x + 9y +9z = 12

-     -       -        -

7y+15z = 0 (4)

Multiply eq(2) with 5 and eq(3) with 4 and subtracting

20x + 15y + 15z = 20

20x + 24y + 28z = 20

-      -         -           -

-9y -13z = 0    (5)

Multiply eq(4) with 9 and eq(5) with 7 and add both equations

63y + 135 z = 0

-63y - 91 z = 0

44z = 0 => z =0

Putting value of z in eq(5)

-9y -13z = 0

-9y -13(0) = 0

-9y = 0

y =0

Putting value of y and z in eq(1)

3x + 4y + 6z = 3

3x + 4(0) +6(0)=3

3x = 3

x =1

So, Value of x = 1, y = 0 and z = 0

Option D is correct.

step-by-step explanation:

we are given

since, we have to isolate x on anyone side

so, we can move all x terms on one side

subtract both sides by 2x

add both sides by 9

divide both sides by 4

we get

6 spoons of sugar for every 21 cups of flour

step-by-step explanation: