Question

what is the solution set for {5(4x+y)+7=-1 and 4(x-5y+3z)=23 and 16(-1+y)=-5(z-5y)

Answer 1

Expanding the parentheses and combining some values:

20x+5y=-8, so 20x+5y+0z=-8,

4x-20y+12z=23,

-5z+9y=-16, so 0x+9y-5z=-16.

Let the determinant D=

| 20   5   0 |

|  4 -20 12 | = 20(100-108)-5(-20-0)+0=-160+100=-60.

|  0   9   -5 |

Determinant D_x=

|  -8   5   0 |

| 23 -20 12 | = -8(100-108)-5(-115+192)+0=64-385=-321.

| -16   9   -5 |

x=D_x/D=-321/-60=5.35.

Determinant D_y=

| 20  -8   0 |

|  4  23 12 | = 20(-115+192)+8(-20-0)+0=1540-160=1380

|  0 -16  -5 |

y=D_y/D=1380/-60=-23.

Determinant D_z=

| 20   5  -8 |

|  4 -20 23 | = 20(320-207)-5(-64-0)-8(36-0)=2260+320-288=2292

|  0   9 -16 |

z=D_z/D=2292/-60=-38.2.

Solution of this system of equations: x=5.35, y=-23, z=-38.2.