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

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

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Answer 1 · 2018-01-19T05:56:04

$x_{1} = 2.5$ $x_1=2.5$
$x_{2} = - 3$ $x_2=-3$
$x_{3} = - 4$ $x_3=-4$

Explanation:

This is a 3-variable system of equations. Values of $x_{1}$ $x_1$ , $x_{2}$ $x_2$ . and $x_{3}$ $x_3$ from these 3 equations can be solved by solving the unknown variables simultaneously. Pair the equations randomly as shown:

Pair 1:

$e q .1 \to 2 x_{1} - 10 x_{2} + 7 x_{3} = 7$ $eq.1->2x_1-10x_2+ 7x_3=7$
$e q .2 \to 6 x_{1} - x_{2} + 5 x_{3} = - 2$ $eq.2->6x_1-x_2+5x_3=-2$

Pair 2:

$e q .2 \to \to 6 x_{1} - x_{2} + 5 x_{3} = - 2$ $eq.2->->6x_1-x_2+5x_3=-2$
$e q .3 \to \to - 4 x_{1} + 8 x_{2} - 3 x_{3} = - 22$ $eq.3->->-4x_1+8x_2-3x_3=-22$

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

$e q .1 \to 2 x_{1} - 10 x_{2} + 7 x_{3} = 7} - 3$ $eq.1->color(red)(2x_1)-color(blue)(10x_2)+color(green)( 7x_3=color(black)(7 "}"-3$
$e q .2 \to 6 x_{1} - x_{2} + 5 x_{3} = - 2$ $eq.2->color(red)(6x_1)-color(blue)(x_2)+color(green)(5x_3)=color(black)(-2)$

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:
$x_1=2.5; x_2=-3; x_3=-4$

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

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

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

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

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

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

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

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Explanation:

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