Question

For the following set of simultaneous equations 1.5x-0.5y=2, 4x+2y+32=9,
7x+y+5=10​

Answer 1

GIVEN :

The set of simultaneous equations are

1.5x-0.5y=2, 4x+2y+3z=9,  and 7x+y+5z=10​

TO FIND :

The solution to the given set of simultaneous equations

SOLUTION :

Given set of simultaneous equations are

$1.5x-0.5y=2\hfill (1)$

$4x+2y+3z=9\hfill (2)$

 and $7x+y+5z=10\hfill (3)$

Now solving the equations (1) ,(2) and (3)

Multiply the equation (3) into 2 we get

$14x+2y+10z=20\hfill (4)$

Now subtracting the equations (2) and (4)

4x+2y+3z=9

14x+2y+10z=20

(-)__(-)__(-)__(-)__

-10x-7z=-11

$10x+7z=11\hfill (5)$

Multiply the equation (3) into 0.5 we get

$3.5x+0.5y+2.5z=5\hfill (6)$

Adding the equations (1) and (6)

1.5x-0.5y=2

3.5x+0.5y+2.5z=5

_______________

$5x+2.5z=7\hfill (7)$

Multiply the equation (7) into 2

$10x+5z=14\hfill (8)$

Subtracting  the equations (5) and (7)

10x+7z=11

10x+5z=14

(-)__(-)__(-)_

   2z=-3

$z=-\frac{3}{2}$

Substitute the value of z in (5)

$10x+7(-\frac{3}{2})=11$

$10x=11+\frac{21}{2}$

$10x=\frac{22+21}{2}$

$x=\frac{43}{20}$

Substitute the value of x and z in equation (2)

$4(\frac{43}{20})+2y+3(\frac{-3}{2})=9$

$2y=9+\frac{9}{2}-\frac{43}{5}$

$2y=\frac{90+45-86}{10}$

$y=\frac{49}{20}$

∴ The solution is ($\frac{43}{20},\frac{49}{20},\frac{-3}{2}$)