Question

The solution of the simultaneous equations (1/2)x + (1/3)y = 2 and x+y=1 is:

Answer 1

Your answer is
y=9
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Answer 2

Answer:

Required solution of given equations are

$x = 10 \\ y = - 9$

Step-by-step explanation:

Given,

$\frac{x}{2} + \frac{y}{3} = 2....(1)$

and $x + y = 1...(2)$

From equation (2),

$x = 1 - y...(3)$

We are putting value of x in equation (1),

$\frac{1 - y}{2} + \frac{y}{3} = 2 \\ \frac{1}{2} - \frac{y}{2} + \frac{y}{3} = 2 \\ \frac{y}{3} - \frac{y}{2} = 2 - \frac{1}{2} \\ \frac{2y - 3y}{6} = \frac{4 - 1}{2} \\ \frac{ - y}{6} = \frac{3}{2} \\ y = - \frac{3}{2} \times 6 \\ y = - 9$

We are putting value of y in equation (3),

$x = 1 - y \\ = 1 - ( - 9) \\ = 1 + 9 \\ = 10$

Required value of x is 10 and required value of y is (-9)

This is a problem of Algebra.

Some important Algebra formulas:

${(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} \\ {(x + y)}^{3} = {x}^{3} + 3 {x}^{2} y + 3x {y}^{2} + {y}^{3} \\ {(x - y)}^{3} = {x}^{3} - 3 {x}^{2} y + 3x {y}^{2} - {y}^{3} \\ {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y) \\ {x}^{3} - {y}^{3} = {(x - y)}^{3} + 3xy(x - y) \\ {x}^{2} - {y}^{2} = (x + y)(x - y) \\ {x}^{2} + {y}^{2} = {(x - y)}^{2} + 2xy \\ {x}^{2} - {y}^{2} = {(x + y)}^{2} - 2xy \\ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} ) \\ {x}^{3} + {y}^{3} = (x - + y)( {x}^{2} - xy + {y}^{2} )$

Know more about Algebra,

1) https://brainly.in/question/13024124

2) https://brainly.in/question/1169549

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