Question

1. Evaluate: 81/36 x square - y square /25​

Answer 1

here is the answer... plz thank me if you think i helped

Answer 2

$\frac{81}{36} x^{2} -\frac{y^{2} }{25} = \frac{1}{10} .(15x+2y).(15x-2y)$.

• A mathematical equation is a formula that uses the equals sign to represent the equality of two expressions. Finding the values of the variables that result in the equality is the first step in solving an equation with variables.
• The unknown variables are also known as the variables for which the equation must be solved, and the unknown variable values that fulfil the equality are known as the equation's solutions. Equations come in two varieties: identities and conditional equations. All possible values of the variables result in an identity. Only certain combinations of the variables' values make a conditional equation true.

Here, according to the given information, we are given that,

$\frac{81}{36} x^{2} -\frac{y^{2} }{25} .$

Now, we know that, the algebraic identity $a^{2} -b^{2}$ is equal to $(a+b)(a-b).$

Applying this, we get,

$\frac{81}{36} x^{2} -\frac{y^{2} }{25} .\\=(\frac{9x}{6} )^{2} -(\frac{y}{5} )^{2} \\=(\frac{9x}{6} +\frac{y}{5} ).(\frac{9x}{6} -\frac{y}{5})\\=(\frac{45x+6y}{30} ).(\frac{45x-6y}{30} )\\=\frac{1}{10} .(15x+2y).(15x-2y)$

Hence, $\frac{81}{36} x^{2} -\frac{y^{2} }{25} = \frac{1}{10} .(15x+2y).(15x-2y)$.

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