Question

How do you solve this system of equations: `2x _ { 1} - 10x _ { 2} + 7x _ { 3} = 7, 6x _ { 1} - x _ { 2} + 5x _ { 3} = - 2 and - 4x _ { 1} + 8x _ { 2} - 3x _ { 3} = - 22`?

Answer 1

`x_1=2.5`
`x_2=-3`
`x_3=-4`

Explanation:

This is a 3-variable system of equations. Values of `x_1`, `x_2`. and `x_3` from these 3 equations can be solved by solving the unknown variables simultaneously. Pair the equations randomly as shown:

Pair 1:

`eq.1->2x_1-10x_2+ 7x_3=7`
`eq.2->6x_1-x_2+5x_3=-2`

Pair 2:

`eq.2->->6x_1-x_2+5x_3=-2`
`eq.3->->-4x_1+8x_2-3x_3=-22`

Now, solve the `1st " pair"` and eliminate one of the variables to form a 2-variable equation that can be considered the `4th` equation. In this case, `x_1` is chosen since the terms are easier to eliminate. Multiply eq.1 by `-3` to eliminate the x terms in both equations; thus,

`eq.1->color(red)(2x_1)-color(blue)(10x_2)+color(green)( 7x_3=color(black)(7 "}"-3`
`eq.2->color(red)(6x_1)-color(blue)(x_2)+color(green)(5x_3)=color(black)(-2)`

These equations formed as shown below.

`eq.1->color(red)(cancel(-6x_1)+color(blue)(30x_2)-color(green)(21x_3)=color(black)(-21)`
`eq.2->color(red)(cancel(6x_1)-color(blue)(x_2)+color(green)(5x_3)=color(black)(-2)`

The derived equation is:

`eq.4->color(blue)(29x_2)-color(green)(16x_3)=-color(black)(23)`

Then, solve the `2nd " pair"` and again eliminate the same variable to form a 2-variable equation that can be considered the `5th` equation. In this case, `x_1` is chosen again to match the derived equation earlier; the `eq.4`. Multiply eq.1 by `-3` to eliminate the x terms in both equations; thus,

`eq.2->color(red)(6x_1)-color(blue)(x_2)+color(green)( 5x_3=color(black)(-2 "} " 2`
`eq.3->color(red)(-4x_1)+color(blue)(8x_2)-color(green)(3x_3)=color(black)(-22 "} " 3`

These equations formed as shown below.

`eq.2->cancel(color(red)(12x_1))-color(blue)(2x_2)+color(green)( 10x_3=color(black)(-4)`
`eq.3->cancel(color(red)(-12x_1))+color(blue)(24x_2)-color(green)(9x_3)=color(black)(-66)`

The derived equation is:

`eq.5->color(blue)(22x_2)+color(green)(x_3)=-color(black)(70)`

Now, solve simultaneously eq.4 and eq.5 to find either of the remaining variables.

`eq.4->color(blue)(29x_2)-color(green)(16x_3)=-color(black)(23)`
`eq.5->color(blue)(22x_2)+color(green)(x_3)=-color(black)(70)`

Multiply eq.5 by 16 to eliminate the term with `x_3`

`eq.4->color(blue)(29x_2)-color(green)(16x_3)=-color(black)(23)`
`eq.5->color(blue)(22x_2)+color(green)(x_3)=-color(black)(70) "} "16`

This can be computed as shown below.

`eq.4->color(blue)(29x_2)-cancel(color(green)(16x_3))=-color(black)(23)`
`eq.5->color(blue)(352x_2)+cancel(color(green)(16x_3))=-color(black)(1120)`

Now, simplify and find the value of `y`. The equation is shown below.

`381x_2=-1143`; divide bothe sides by 381 to isolate `x`
`(cancel(381)x_2)/cancel(381)=-1143/382`
`color(red)(x_2=-3`

Knowing the value of `x_2`, helps determine the value of the other variable. Use equation 5 to find the value of `x_3`. Substitute value `x_2=-3` to the equation; i.e,

`22x_2+x_3=-70`
`22(-3)+x_3=-70`
`-66+x_3=-70`
`x_3=-70+66`
`color(red)(x_3=-4)`

Finally, find the value of `x_1`. Use eq.1 to get its value by the following solving processes involved.

where:

`x_2=-3`
`x_3=-4`

`2x_1-10x_2+ 7x_3=7`
`2x_1-10(-3)+7(-4)=7`
`2x_1+30-28=7`
`2x_1+2=7`
`2x_1=7-2`
`2x_1=5`
`x_1=5/2`
`color(red)(x_1=2.5)`

Checking:
`x_1=2.5; x_2=-3; x_3=-4`

`eq.1->2x_1-10x_2+ 7x_3=7`

`eq.1->2(2.5)-10(-3)+7(-4)=7`
`eq.1->5+30-28=7`
`eq.1->5+2=7`
`eq.1->7=7`

`eq.2->6x_1-x_2+5x_3=-2`

`eq.2->6(2.5)-(-3)+5(-4)=-2`
`eq.2->15+3-20=-2`
`eq.2->18-20=-2`
`eq.2->-2=-2`

`eq.3->->-4x_1+8x_2-3x_3=-22`

`eq.3->-4(2.5)+8(-3)-3(-4)=-22`
`eq.3->-10-24+12=-22`
`eq.3->-34+12=-22`
`eq.3->-22=-22`