Question

[If the ordered pairs (a, - 1) and (5, b) belong to \ (x,y)/y=2x+3\ is, then the values of a and b are​

Answer 1

Answer:

put (a,-1) and (5,b) in the equation: y=2x+3

when we put (a,-1),

y=2x+3

(-1)=2(a)+3

2a= -1-3 = -4

a=-2

when we put (5,b),

b=2(5)+3 = 10+3

b=13

a=-2 and b=13

Answer 2

Answer:

The values of a is (-2) and b is 13.

Step-by-step explanation:

Given,

$y = 2x + 3...(1)$

It is also given that, (a, - 1) and (5, b) belong to (x,y).

So we are putting (a,-1) in equation (1).

$- 1 = 2 \times a + 3 \\ - 1 = 2a + 3 \\ 2a = - 1 - 3 \\ 2a = - 4 \\ a = \frac{ - 4}{2} \\ a = - 2$

and we are putting (5,b) in equation (1),

$b = 2 \times 5 + 3 \\ b = 10 + 3 \\ b = 13$

Therefore, required value of a is (-2) and 13.

This is a problem of Algebra.

Some important formulas of Algebra :

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

Two more important Algebra problem:

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

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

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