Question

Solve the system by substitution.or522y + 3z = 62x - 4y - 3z = 14-x+6y - 8z = 12

Answer 1

Given a system of equations are

`5x-2y+3z=6\text{ take it as equation (1)}`
`2x-4y-3z=14\text{ take it as equation (2)}`
`x+6y-8x=12\text{ take it as equaiton (3)}`

Adding equation (1) and equation (2), we get

`(5x-2y+3z)+(2x-4y-3z)=6+14`

`5x+2x_{}-2y-4y+3z-3z=20`

`7x_{}-6y=20\text{ take it as equation (4)}`

Multiplying equation (2) by 8, we get

`8*2x-8*4y-8*3z=8*14`

`16x-32y-24z=112\text{ take it as equation (5)}`

Multiplying equation (3) by (-3), we get

`(-3)x+(-3)6y-(-3)8x=(-3)12\text{ }`

`3x-18y+24x=-36\text{ take it as equation (6)}`

Adding equation (5) and equation (6), we get

`(16x-32y-24z)+(3x-18y+24x)=112-36`

`16x+3x-32y-18y-24z+24x=76`

`19x-50y=76`

Adding 50 on both sides, we get

`19x-50y+50y=76+50y`

`19x=76+50y`

Dividing by 19. we get

`x=(76+50y)/(19)`

`\text{ Substitute }x=(76+50y)/(19)\text{ in equation (4) as follows}`

`7((76+50y)/(19)_{})-6y=20\text{ }`

`((7*76+7*50y)/(19)_{})-(19*6y)/(19)=20\text{ }`

`(532+350y-114y)/(19)=20\text{ }`

`532+236y=20*19`

`236y=380-532`

`y=-(152)/(236)`
`y=-0.644`

`\text{ Substitute y=-0.644 in equation }x=(76+50y)/(19)\text{as follows}`
`\text{x}=(76+50(-0.644))/(19)`

`x=2.301`

Substitute x=2.301 and y=-0.644 in equation (1), we get

`5(2.301)-2(-0.644)+3z=6`

`12.838+3z=6`
`z=(6-12.828)/(3)`
`z=-2.279`

Hence the solutions are

`x=2.301`
`y=-0.644`

`z=-2.279`