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# solve following

Asked by abhinavsinha 24th May 2010, 9:06 PM

Dear Student,

1.  Hence, the final solution is             x = 4 , y = 5

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2.

x – y + z = 4                                        ………(i)

x 2y 2z = 9                                    ………(ii)

2x + y + 3z = 1                                    ………(iii)

Here, we have linear equations in three variables.

First of all, we will find a system of linear equations in two variables by eliminating one variable from these three equations

Equation (i) + Equation (iii) gives

x + 2x + z + 3z = 4 + 1

i.e.

3x + 4z = 5                   ……(iv)

Also 2 × {Equation (i)} – Equation (ii) gives

2(x – y + z) – (x 2y 2z) = 8 – 9

i.e.

x + 4z = – 1  ……… (v)

Now, equations (iv) and (v) represent a system of linear equations in two variables x and z

Now, we solve these two equations by substitution method

From equation (v), x = –1 – 4z

Substituting this value of x in (iv), we get

3(–1 – 4z)  + 4z = 5

– 3 – 12z + 4z = 5

– 8z = 8

So, z = – 1

Substituting for z in (v), we get x + 4(–1) = – 1

So, x = 3

Substituting for x and z in (i), we get

3 – y + (–1) = 4

y = – 2

Hence, the solution is

x = 3

y = – 2

z = – 1

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3.  Or

3x + y = 4        ………………(v)   and

3x – y = 2        ………………(vi)

Adding (v) and (vi), we get 6x = 6 which means x = 1

And substituting this value of x in (v), we get  3 + y = 4 which means y = 1

Hence, the final solution is  x = 1, y = 1

Regards

Team

Topper Learning

Answered by Expert 25th May 2010, 2:30 PM
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