Mathematics, 20.09.2020 15:01 tina7659

Identify the number of solutions of the system of linear equations. 5x+y−z=6 x+y+z=2 12x+4y=10

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Answer from: Barbiealice8362

x = $\frac{19}{12}$ y = $(-\frac{9}{4} )$ z = $\frac{8}{3}$ One solution for each variable.

Step-by-step explanation:

5x + y − z = 6

x + y + z = 2

12x + 4y = 10

The first thing we need to do is solve for x in the 3rd equation because it inly have 2 variables, x and y.

12x + 4y = 10 Subtract 4y from each side

12x + 4y - 4y = 10 - 4y

12x = 10 - 4y Pull 2 out on the right side

12x = 2(5 - 2y) Divide each side by 12

12x/12 = 2(5 - 2y)/12

x = 2(5 - 2y)/12

x = (5 - 2y)/6

Now we plug in our x value into the 2nd equation and solve for z

x + y + z = 2

$\frac{5-2y}{6}$ + y + z = 2 Multiply each side by 6

6( $\frac{5-2y}{6}$ + y + z) = 2 * 6

6( $\frac{5-2y}{6}$ + y + z) = 12

5 - 2y + 6y + 6z = 12 Combine like terms

5 + 4y + 6z = 12 Subtract 5 from each side

5 - 5 + 4y + 6z = 12 - 5

4y + 6z = 7 Subtract 4y from each side

4y - 4y + 6z = 7 - 4y

6z = 7 - 4y Divide each side by 6

6z/6 = (7 - 4y)/6

z = (7 - 4y)/6

Now we solved for z and x, so in the 1st equation we plug in x and z.

5x + y − z = 6

5( $\frac{5-2y}{6}$ ) + y - $\frac{7-4y}{6}$ = 6 Multiply each side by 6

6*(5( $\frac{5-2y}{6}$ ) ) + 6y - 6( $\frac{7-4y}{6}$ ) = 6*6

6*(5( $\frac{5-2y}{6}$ ) ) + 6y - 6( $\frac{7-4y}{6}$ ) = 36

5(5 - 2y) + 6y - 7 - 4y = 36

25 - 10y + 6y - 7 - 4y = 36 Rearrange to make it easier to combine terms.

25 - 7 - 10y + 6y - 4y = 36

18 - 8y = 36 Subtract 18 from each side.

18 - 18 - 8y = 36 - 18

- 8y = 36 - 18

- 8y = 18 Divide each side by -8

- 8y/-8 = 18/- 8

y = 18/- 8

y = - 9/4

Now we plug our answer for y back into the 3rd equation and solve for the value of x.

12x + 4y = 10

12x + 4 $(-\frac{9}{4} )$ = 10

12x - 9 = 10 Add 9 to each side

12x - 9 + 9 = 10 + 9

12x = 10 + 9

12x = 19 Divide each side by 12

12x/12 = 19/12

x = 19/12

Now we have a value for x and y so plug these into the 2nd equation to sovle for z.

x + y + z = 2

$\frac{19}{12}$ + $(-\frac{9}{4} )$ + z = 2 We need to find the common denominator in order to add.

$(-\frac{9}{4} )$ * $\frac{3}{3}$ = $-\frac{27}{12}$

$\frac{19}{12}$ $-\frac{27}{12}$ + z = 2

$-\frac{8}{12}$ + z = 2 Add $-\frac{8}{12}$ to each side

$-\frac{8}{12}$ $+ \frac{8}{12}$ + z = 2 $+ \frac{8}{12}$

z = 2 $+ \frac{8}{12}$ Reduce $+ \frac{8}{12}$ to $\frac{2}{3}$

z = 2 + $\frac{2}{3}$ To add find a common denominator.

$2 * \frac{3}{3} = \frac{6}{3}$

z = $\frac{6}{3}$ + $\frac{2}{3}$

z = $\frac{8}{3}$

So there is 1 solution for each variable.

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