Question

Is there a value of r so that x=1,y=2, z=r is a solution of the following linear system.if there is, find it . 2x+3y-z =11 , x-y+2z=-7, 4X+y-2z=12​

Answer 1

Step-by-step explanation:

2x+3y-z=11

2(1) +3(2)-r=11

2+6-r=11

r=-3

x-y+2z=-7

1-2+2(r)=-7

-2+2(r)= -7-1

2(r) = -8+2

2(r)= -6

r= -3

4x+y-2z=12

4(1)+2-2(r) = 12

2-2(r)=12-4

-2(r)=8-2

r=6/-2

r=-3

Answer 2

Answer: To find if there is a value of r so that x=1, y=2 and z=r is a solution of the given linear system, we substitute the values of x and y in the three equations and solve for z.

Step-by-step explanation:

Using the values of x=1 and y=2, the three equations become:

2(1) + 3(2) - z = 11

1 - 2r = -7

4(1) + 2 - 2r = 12

Simplifying the second equation, we get:

1 - 2r = -7

2r = 8

r = 4

Substituting the value of r=4 in the three equations, we get:

2(1) + 3(2) - 4 = 11

1 - 2(4) = -7

4(1) + 2 - 2(4) = 12

Thus, the value of r=4 satisfies the given linear system, and x=1, y=2, and z=4 is a solution.

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