Respuesta :

altavistard
4x-5y=5 -0.8x+0.10y=0.10 should be presented like this:

4x-5y=5
-0.8x+0.10y=0.10

Mult. the 2nd eqn by 100 to eliminate the fractions:

-80x + 10y = 10

Mult all terms of the first equation by 20:  80x -100y = 100

Now combine the two equations:

-80x + 10y = 10
 80x -100y = 100
-------------------------           -11
          -90y = 110, or y = --------
                                            9

This is a first-order linear equation, so there is only one solution.  having found y = -11/9, you must now find x by subbing -11/9 for y in the first given equation.

The system of equations 4x-5y=5 and -0.8x+0.10y=0.10 have no solutions.

How do we find the number of solution of linear equation?

Consider the pair of linear equations in two variables x and y.

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

Here a1, b1, c1, a2, b2, c2 are all real numbers.

If [tex]\frac{a1}{a2} \neq \frac{b1}{b2}[/tex], then there will be a unique solution.

If [tex]\frac{a1}{a2} = \frac{b1}{b2}=\frac{c1}{c2}[/tex], then there will be infinitely many solutions.

If [tex]\frac{a1}{a2} = \frac{b1}{b2} \neq \frac{c1}{c2}[/tex], then there will be no solution.

Standard equation of line: Ax + By + C = 0

4x - 5y - 5 = 0

a1 = 4

b1 = -5

c1 = -5

- 0.8x + 0.10y - 0.10 = 0

a2 = -0.8

b2 = 0.1

c2 = -0.1

[tex]\frac{a1}{a2} = \frac{4}{-0.8} = -5\\ \\ \frac{b1}{b2} = \frac{-5}{0.1} = -5\\\\ \frac{c1}{c2} = \frac{-5}{-0.1} = 5\\[/tex]

[tex]\frac{a1}{a2} = \frac{b1}{b2} \neq \frac{c1}{c2}[/tex]

This implies that the given system of equation has no solution.

Learn more about number of solution of linear equation here

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