-6x-14y=-20 -3x-7y=-10

1 answer:

6 0

ANSWER:

x = 10 / 3

y = 0

STEP-BY-STEP EXPLANATION:

We will be using simultaneous equations to solve this problem. Let's first establish the two equations which we will be using.

Equation No. 1 -

- 6x - 14y = - 20

Equation No. 2 -

- 3x - 7y = - 10

First, we will make ( x ) the subject in the first equation and simplify accordingly.

Equation No. 1 -

- 6x - 14y = - 20

- 6x = - 20 + 14y

x = ( - 20 + 14y ) / - 6

x = ( - 10 + 7y ) / - 3

From this, we will make ( y ) the subject in the second equation and substitute the value of ( x ) from the first equation into the second equation to solve for ( y ) accordingly.

Equation No. 2 -

- 3x - 7y = - 10

- 7y = - 10 + 3x

- 7y = - 10 + 3 [ ( - 10 + 7y ) / - 3 ]

- 7y = - 10 + [ ( - 30 + 21y ) / - 3 ]

- 7y = - 10 + ( 10 - 7y )

- 7y = - 7y

- 7y + 7y = 0

0y = 0

y = 0

Using this, we will substitute the value of ( y ) from the second equation into the first equation to solve for ( x ) accordingly.

x = ( - 10 + 7y ) / - 3

x = [ - 10 + 7 ( 0 ) ] / - 3

x = [ - 10 + 0 ] / - 3

x = - 10 / - 3

x = 10 / 3

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if we have a quadratic equation, $ax^2 +bx +c = 0$ , where $a$ , $b$ and $c$ are real numbers and $a \neq 0$ , $x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$ .

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