System Of linear Equations-Exercise solved

1.

Please copy the question from book.

a.

Soln:

Or, x + 2y = 5 ….(i)

Or, 5x – 3y = -1 …(ii)

Multiplying equation (i) by 5 and then subtracting from (ii).

Or, -13y = -26. ….(iii)

Now, we have, the following equations

Or, x + 2y = 5 …(i)

Or, -13y = -26 …(iii)

From equation (iii), -13y = -26.

Or, y = 2.

Using y = 2 in (i), we have,

Or, x + 2 * 2 = 5.

Or, x = 1.

So, x = 1, y = 2.

b.

Soln:

Or, 3x – 2y = 5….(i)

Or, 4x – y = 10 …(ii)

Multiplying equation (i) by 43 and then subtracting from (ii),

Or, 53y = 103.

We have the following equations,

or, 3x – 2y = 5….(i)

or, 53y = 103 …(iii)

From equation (iii),

Or, 53y = 103.

So, y = 2.

Using y = 2 in (i) we have,

Or, 3x – 2 * 2 = 5.

Or, 3x = 9.

Or, x = 3.

So, x = 3, y = 2.

c.

Soln:

Or, 2x + 7y = 3 …(i)

Or, 6x – 5y = -17 …(ii)

Multiplying equation (i) by 3 and the subtracting from (ii),

Or, -26y = -26.

Now, we have the following equations,

Or, 2x + 7y = 3….(i)

Or, -26y = -26 ….(iii),

From equation (iii), we have,

Or, -26y = -26.

So, y = 1.

Using y = 1in (i),

Or, 2x + 7 * 1 = 3.

Or, 2x = -4.

So, x = -2.

So, x = -2, y = 1.

d.

Soln:

5x – 8y = 28 …(i)

Or, 3x + 7y = 5 ….(ii)

Multiplying equation (I) by 35 and then subtracting from (ii),

Or, 59y5 = −595

Now, we have the following equations:

Or, 5x – 8y = 28…..(i)

Or, 59y5 = −595…..(iii)

From equation (iii),

Or, 59y5 = −595

So, y = -1.

Using y = -1 in (i), we have,

Or, 5x – 8 * (-1) = 28.

Or, 5x = 20

So, x = 4.

So, x = 4, y = -1.

2.

Please copy the question from book.

a.

Soln:

Or, x + 3y – 2z = 0 …(i)

Or, 2x – 3y + z = 1 …(ii)

Or, 4x – 3y + z = 3 …(iii)

Multiplying equation (1) by 2 and subtracting (ii) from (i),

Or, 9y – 5z = -1 …(iv)

Multiplying equation (i) by 4 and subtracting (iii) from (i),

Or, 5y – 3z = 01 …(v)

Multiplying (iv) by 5 and (v) by 9 subtracting (iv) from (v),

Or, 2z = 4.

Now, we have the following three equations,

Or, x + 3y – 2z = 0 …(i)

Or, 5y – 3z = - 1 …(iv)

Or, 2z = 4 …(vi)

From equation (vi), we have z = 2.

Using of z = 2 in (iv), we have, y = 1.

Using y = 1 and z = 2 in (i), we have x = 1,

So, x = 1, y = 1 , z = 2.

b.

Soln:

Or, x + 3y – z = -2….(i)

Or, 3x + 2y – z = 3 …(ii)

Or, -6x – 4y – 2z = 18 …(iii)

Multiplying equation (i) by 3 and subtracting (ii),

Or, -7y – 2z = 9 …(iv)

Multiplying equation (iv) by 2 and subtracting (iv),

Or, -4z = 24….(v)

Now, we have the following three equations:

Or, x + 3y – z = -2…(i)

Or, -7y + 2z = 9 ….(iv)

Or, -4z = 24 ….(v)

From equation (vi), we have z = -6,

Substituting the value in z in (ii),

Or, -7y + 2 * (-6) = 9.

Or, -7y = 9 + 12 = 21.

So, y = -3.

Substituting the values of y and z in (i),

Or, x + 3 * (-3) – (-6) = -2

Or, x – 9 + 6 = -2.

So, x = 3 – 2 = 1

So, the solution is x = 1, y = -4, z= -6.

So, the solution is x = 1, y = -3 and z= -6.

c.

Soln:

Or, x + 3y – 2z = 5 …(i)

Or, 3x + 5y + 6z = 7…(ii)

Or, 2x + 4y + 3z = 8 …(iii)

Multiplying equation (i) by 3 and subtracting (ii) from (i),

Or, 4y – 12z = 8 …(iv)

Multiplying (i) by 2 and subtracting (iii) from (i),

Or, 2y – 7z = 2 ….(v)

Multiplying equation (v) by 2 and subtracting from (iv),

Or, 2z = 4 ….(vi).

Now, we have the three equations,

Or, x + 3y – 2z = 5 ….(i)

Or, 4y – 12z = 8 …(iv)

Or, 2z = 4 ….(vi)

From equation (vi), we have z = 2,

Use of z = 2 in equation (iv), we have, y = 8.

Using y = 8 and z = 2 in (i), we have, x= -15.

So, x = -15, y = 8, z = 2.

d.

Soln:

Or, 2x – 3y + 3x = 27 …(i)

Or, 4x + y – 2z = 0 …(ii)

Or, -6x – 4y + 2z = 0 …(iii)

Multiplying equation (i) by 2 and subtracting (ii) from (i),

Or, 7y – 8z = -54 …(iv)

Multiplying equation (i) by 3 and adding with (iii),

Or, -13y + 11z = 81 ….(v)

Multiplying equation (iv) by 13 and (v) by 7 and adding,

Or, -27z = -135 …(iv)

Now, we have three equations,

Or, 2x – 3y + 3z = 27 …(i)

Or, 7y – 8z = -54….(iv)

Or, -27z = -135.

From equation (vi), z = 5.

Use of z = 5 in equation (iv) we have, y = - 2.

Using y = -2 and z = 5 in equation (i), we have, x = 3.

So, x = 3 , y = - 2, z = 5.

e.

Soln:

3x + 2y – z = 1 …(i)

Or, 2x – 2y + 4z = -2 …(ii)

Or, -x + 12y – z = 0 …(iii)

Multiplying equation (I) by 2 and 3 subtracting (ii) from (i),

Or, 10y – 14z = 8 …(iv)

Multiplying equation (iii) by 3 and adding with (i),

Or, 72y – 4z = 1….(v)

Multiplying equation (Iv) 72 and (v) by 10 and subtracting (v) from (iv),

Or, - 9z = 18 ….(vi)

Now, we have three equations,

Or, 3x + 2y – z = 1. …(i)

Or, 10y – 14z = 8 …(iv)

Or, -9z = 18 …(vi)

From equation (vi), we have, z = -2.

Use of z = -2 in (iv), we have, y = -2.

Using y = -2 and z = -2 in equation (i), we have, x = 1.

So, x = 1, y = -2, z = -2.

f.

Soln:

The given system of equations is:

Or, 3x1 + 6x2 + x3 = 16 …(i)

Or, 2x1 + 4x2 + 3x3 = 13 ….9ii)

Or, x1 + 3x2 + 2x3 = 9 …9 (iii)

Multiplying equation (i) by 23$\frac{2}{3}$ and then subtracting from (ii)

Or, 73x3 = 73

Again, multiplying equation (i) by 13and then subtracting from (iii),

Or, x2 + 53x3 = 113 …(v)

Now, we have the following equation:

Or, 3x1 + 6x2 + x3 = 16 ….(i)

Or, x2 + 53.x3 = 113 …(v)

Or, 73x3 = 73 …(iv)

From equation (iv),

Or, 73x3 = 73

Or, x3 = 1,

Using x3 = 1 in (v),

Or, x2 + 53 = 113.

Or, x2 = 113−53= 2.

Using x2 = 2 and x3 = 1 in (i),

Or, 3x1 + 6 * 2 + 1 = 16.

Or, 3x1 = 3

So, x1 = 1

So, x1 = 1, x2 = 2, x3 = 1.

g.

Soln:

Or, x1 – 2x2 + 3x3 = 10 …(i)

Or, 2x1 + 3x2 – 2x3 = 1 …(ii)

Or, -x1 – 2x2 + 4x3 = 13 ….(iii)

Multiplying equation (i) by 2 and subtracting from (ii),

Or, 7x2 – 8x3 = -19 ….(iv)

Adding (i) and (iii),

Or, - 4x2 + 7x3 = 23 ….(v).

Multiplying equation (iv) by 4 and (v) by 7 and adding both,

Now, we have three equations,

Or, x1 – 2x2 + 3x3 = 10 ….(i)

Or, 7x2 – 8x3 = 19 ….(iv)

Or, 17x3 = 85…..(vi)

From (vi), we have, x3 = 5.

Use of x3 = 5 in (iv), we have x2 = 3.

Using x2 = 3, x3 = 5 in equation (i), we have, x1 = 1.

So, x1 = 1,x2 = 3,x3 = 5.

h.

Soln:

Or, 3x1 – x2 + x3 = 2 …(i)

Or, - 15x1 + 6x2 – 5x3 = 5 …(ii)

Or, 5x1 – 2x2 + 2x3 = 1 …(iii),

Multiplying equation (i) by 5 and adding with (ii),

Or, x2 = 15 ….(iv)

Multiplying equation (i) by 5 and (iii) by 3 and subtracting (iii) from (i),

Or, x2 + x3 = 7….(v)

Now, we have three equations,

Or, 3x1 – x2 + x3 = 2 …(i)

Or, x2 = 15 …(iv)

Or, x2 – x3 = 7 …(v)

From (iv), we have, x2 = 15.

Use of x2 = 5 in (v), we have, x3 = 8.

Using x2 = 15, x3 = 8 in equation (i), we have, x1 = 3.

So, x1 = 3, x2 = 15, x3 = 8.

3.

Please copy the question from book.

a.

Soln:

We have,

Or, x1 – x2 + x3 = 1….(i)

Or, 3x1 + x2 + 5x3 = 11 …(ii)

Or, 4x1 + 2x2 + 7x3= 16 …(iii),

Multiplying equation (i) by 3 and subtracting (ii) from (i),

Or, -4x2 – 2x3 = -8 ….(iv)

Multiplying equation (i) by 4 and subtracting (iii) from (i),

Or, -6x2 – 3x3 = -12…..(v)

Multiplying equation (iv) by 3 and (v) by 2 and subtracting (v) from (iv),

Or, 0 = 0

Now, we have the following three equations,

Or, x1 – x2 + x3 = 1….(i)

Or, -4x2 – 2x3 = -8…..(iv)

Or, 0.x3 = 0 ….(vi)

This last equation is true for all values of x3. We can assign any number of values to this free variable. Hence, we get an infinitely many solution in such a case. Thus, is x3 = k, then:

Or, x2 = 4−k2 and x1 = 3 – k.

So, x1 = 3 – k,x2 = 4−k2 and x3 = k.

b.

Soln:

We have,

Or, x + 3y + 4z = 8 …(i)

Or, 2x + y + 2z = 5 …(ii)

Or, 5x + 2z = 7 ….(iii).

Multiplying equation (i) by 2 and subtracting (ii) from (i),

Or, 5y + 6z = 11 ….(iv)

Multiplying equation (i) by 5 and subtracting (iii) and (i),

Or, 15y + 18z = 33 ….(v)

Multiplying equation (iv) by 3 and subtracting (v),

0 = 0.

So, we have the following three equations:

Or, x + 3y + 4z = 8 ….(i)

Or, 5y + 6z = 11 ….(iv)

Or, 0.z = 0 .,..(vi)

The lest equation is true for all values of z. We can assign any number of values to this free variable. Hence, we get an infinitely many solution in this case.

If we suppose z = k, then.

Or, y = 11−6k5

Or, x = 7−2k5 and z = k.

c.

Soln:

We have,

Or, x1 + x2 + x3 = -3 …(i)

Or, 3x1 + x2 – 2x3 = -2 …(ii)

Or, 2x1 + 4x2 + 7x3 = 7 …(iii)

Multiplying equation (I) by 3 and subtracting (ii) from (i),

Or, 2x2 + 5x3 = 7 ….(iv)

Multiplying equation (i) by 2 and subtracting (iii) from (i),

Or, -2x2 – 5x3 = 13 …(v)

Adding equation (iv) and (v),

0 = 20.

Now, we have the following three equations.

Or, x1 + x2 + x3 = -3….(i)

Or, 2x2 + 5x3 = 7 ….(iv)

Or, 0.x3 = 20 ….(vi)

Hence, no value of x3 satisfies the equation (vi),

So, the system has no solution.

d.

Soln:

Or, x1 + 2x2 + 3x3 = 4 …(i)

Or, 4x1 + 5x2 + 6x3 = -7 ….(ii)

Or, 7x1 + 8x2 + 9x3 = 10 …… (iii)

Multiplying equation (i) by 4 and then subtracting from (ii),

Or, -3x1 – 6x3 = -23 ….(iv).

Again, multiplying equation (i) by 7 and then subtracting from (iii),

Or, - 6x2 – 12x3 = -18….(v)

Multiplying equation (iv) by 2 and then subtracting from (v),

0 = 28 …(vi)

Now, we have the following equations,

Or, x1 + 2x2 + 3x3 = 4….(i)

Or, - 3x2 – 6x3 = -23 …(iv)

Or, 0.x3 = 28 ….(vi).

No value of x3 satisfies equation (vi)

So, the system has no solution,

That, is the system of equation is inconsistent.

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