Solve the following system of equations using any method 2x+4y+1z=−35 3x+7y+7z=−34 2x+10y+6z=−64

values of x,y and z are x = -2, y= -9 and z=5

Step-by-step explanation:

2x+4y+1z=−35    eq(1)

3x+7y+7z=−34    eq(2)

2x+10y+6z=−64  eq(3)

We can solve using elimination method

Subtracting eq (1) from eq(3)

2x + 10y +6z = -64

2x +4y +1z = -35

6y + 5z = -29     eq(3)

Multiplying eq(2) with 2 and eq(3) with 3 and subtracting

6x + 14y +14z = -68

6x + 30y +18z = -192

-     -         -          +

-16y -4z = 124     eq(4)

Multiply eq(3) with 4 and eq(4) with 5 and add both equations

24y + 20z = -116

-80y - 20z = 620

-56y = 504

y = -504/56

y= -9

Putting value of y in equation(3)

6y + 5z = -29

6(-9) + 5z = -29

-54 + 5z = -29

5z = -29+54

5z = 25

z = 25/5

z =5

Now, putting value of y and z in eq(1)

2x + 4y +1z = -35

2x + 4(-9) +1(5) = -35

2x -36+5 = -35

2x -31 = -35

2x = -35+31

2x = -4

x= -4/2

x=-2

So, values of x,y and z are x = -2, y= -9 and z=5

hope it

step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6)   ·x^(1/6) = x^(4/6)   since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)