Question

10. If x²-1+y^m-4 = 0 is a linear equation, which of these is also a linear equation? solve and explain​

Answer 1

Correct question is "If ${x}^{2n} - 1 + {y}^{m - 4 } = 0$ is a linear equation, which of these is also a linear equation?

Option 1.${x}^{n } + {y}^{m } = 0$

Option 2.${x}^{ \frac{n}{5} } + {y}^{ \frac{m}{5} } = 0$

Option 3.${x}^{ n + \frac{1}{2} } + {y}^{ m + 4 } = 0$

Option 4.${x}^{ \frac{1}{2n} } + {y}^{ \frac{m}{5} } = 0$"

Answer:

${x}^{ \frac{1}{2n} } + {y}^{ \frac{m}{5} } = 0$ is a linear equation.

Option four is the correct answer.

Explanation:

Given,

${x}^{2n} - 1 + {y}^{m - 4} = 0......(1)$

It is given that this is a linear equation.

We know every variable of a linear equation has power one.

So we can say,

equation (1) is equal to

${x}^{1} - 1 + {y}^{1} = 0.......(2)$

So from equation (1) and (2),

$2n = 1 \\ n = \frac{1}{2}$ and $m - 4 = 1 \\ m = 1 + 4 \\ m = 5$

By option test we can find the required answer,

Option -1:

${x}^{n} + {y}^{m} = 0 \\ {x}^{ \frac{1}{2} } + {y}^{5} = 0$

By linear equation rule,this is not a linear equation.

Option -2:

${x}^{ \frac{n}{5} } + {y}^{ \frac{m}{5} } = 0 \\ {x}^{ \frac{ \frac{1}{2} }{5} } + {y}^{ \frac{5}{5} } = 0 \\ {x}^{ \frac{1}{10} } + {y}^{1} = 0$

By linear equation rule,this is not a linear equation.

Option -3:

${x}^{n + \frac{1}{2} } + {y}^{m + 4} = 0 \\ {x}^{ \frac{1}{2} + \frac{1}{2} } + {y}^{5 + 4} = 0 \\ {x}^{1} + {y}^{9} = 0$

By linear equation rule,this is not a linear equation.

Option -4:

${x}^{ \frac{1}{2n} } + {y}^{ \frac{m}{5} } = 0 \\ {x}^{ \frac{1}{ \frac{2}{2} } } + {y}^{ \frac{5}{5} } = 0 \\ {x}^{1} + {y}^{1} = 0$

By linear equation rule,this is a linear equation.

This is a problem of Algebra.

Some important Algebra formulas.

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab − b²

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)³ − 3ab(a + b)

a³ - b³ = (a -b)³ + 3ab(a - b)

a² − b² = (a + b)(a − b)

a² + b² = (a + b)² − 2ab

a² + b² = (a − b)² + 2ab

a³ − b³ = (a − b)(a² + ab + b²)

a³ + b³ = (a + b)(a² − ab + b²)

Know more about Algebra,

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

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

Answer 2

Correct Question

If $x^{2n} -1+y^{m-4} = 0$ is a linear equation, which of these is also a linear equation? solve and explain​.

Option 1. $x^{n} +y^{m} =1$

Option 2. $x^{\frac{n}{5} } +y^{\frac{m}{5} } =1$

Option 3. $x^{ n+\frac{1}{2} } +y^{m+4 } =1$

Option 4. $x^{\frac{1}{2n} } +y^{\frac{m}{5} } =1$

Answer:

A linear equation is $x^{\frac{1}{2n} } +y^{\frac{m}{5} } =1$ and option 4 is correct.

Explanation:

Given:

$x^{2n} -1+y^{m-4} = 0$ is a linear equation.

To Find:

Which is the linear equation from the given options.

Concept Used:

Linear equation $ax + by = c,$, where a, b, and c are real numbers and a ≠ 0 and b≠ 0  has two variables x and y.Hence it is a linear equation with two variables.

These are also called as first‐degree equations, because the highest exponent on the variable is 1.

Solution:

As given, $x^{2n} -1+y^{m-4} = 0$ is a linear equation.

We know that  a linear equation has highest exponent on the variable is 1.

Therefore,  in given equation, exponent on the variable $x$ will be  equal to 1.

$2n=1\\n=\frac{1}{2}$

Similarly.In given equation,  the value of exponent on the variable y will be  equal to  1.

$m-4=1\\m=5$

Option 1. $x^{n} +y^{m} =1$.

$x^{n} +y^{m} =1$

Putting the value of m and n, we get.

$x^{\frac{1}{2} } +y^{5} =1$

In equation $x^{\frac{1}{2} } +y^{5} =1$,  the value of exponent on the variables $x \ and \ y$      is not equal to 1. Therefore, $x^{n} +y^{m} =1$ is not a linear equation.

Option 2. $x^{\frac{n}{5} } +y^{\frac{m}{5} } =1$

$x^{\frac{n}{5} } +y^{\frac{m}{5} } =1$

Putting the value of m and n, we get.

$x^{\frac{(\frac{1}{2}) }{5} } +y^{\frac{5 }{5} } =1$

$x^{\frac{1}{10} } +y } =1$

In equation $x^{\frac{1}{10} } +y } =1$ ,  the value of exponent on the variables $x \ and \ y$      is not equal to 1. Therefore, $x^{\frac{n}{5} } +y^{\frac{m}{5} } =1$ is not a  linear equation.

Option 3. $x^{ n+\frac{1}{2} } +y^{m+4 } =1$.

$x^{ n+\frac{1}{2} } +y^{m+4 } =1$

Putting the value of m and n, we get.

$x^{ \frac{1}{2} +\frac{1}{2} } +y^{5+4 } =1$

$x^1 +y^{9 } =1$

$x +y^{9 } =1$

In equation $x +y^{9 } =1$,  the value of exponent on the variables $x \ and \ y$      is not equal to 1. Therefore, $x^{ n+\frac{1}{2} } +y^{m+4 } =1$ is not a linear equation.

Option 4. $x^{\frac{1}{2n} } +y^{\frac{m}{5} } =1$.

$x^{\frac{1}{2n} } +y^{\frac{m}{5} } =1$

Putting the value of m and n, we get.

$x^{\frac{1}{2\times(\frac{1}{2}) } } +y^{\frac{5}{5} } =1$

$x^{\frac{1}{1} } +y^{\frac{5}{5} } =1$

$x+y=1$

In equation $x+y=1$ ,  the value of exponent on the variables $x \ and \ y$      is  equal to 1. Therefore, $x^{\frac{1}{2n} } +y^{\frac{m}{5} } =1$ is a linear equation.

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