Question

Find the value of the expression $36{{x}^{2}}+25{{y}^{2}}+60xy$ when $x=\dfrac{1}{5}\text{ and }y=\dfrac{4}{3}$ .

Answer 1

Hint: To find the value of the expression $36{{x}^{2}}+25{{y}^{2}}+60xy$ when $x=\dfrac{1}{5}\text{ and }y=\dfrac{4}{3}$ , Let us substitute $x=\dfrac{1}{5}\text{ and }y=\dfrac{4}{3}$ in $36{{x}^{2}}+25{{y}^{2}}+60xy$ . We will get $36{{\left( \dfrac{1}{5} \right)}^{2}}+25{{\left( \dfrac{4}{3} \right)}^{2}}+60\times \dfrac{1}{5}\times \dfrac{4}{3}$ . Solving this, we will get $\dfrac{36}{25}+\dfrac{400}{9}+16$ . Take the LCM of the denominators and solve further to get the required solution.

Complete step by step answer:
We have to find the value of the expression $36{{x}^{2}}+25{{y}^{2}}+60xy$ when $x=\dfrac{1}{5}\text{ and }y=\dfrac{4}{3}$ .
Let us substitute $x=\dfrac{1}{5}\text{ and }y=\dfrac{4}{3}$ in the expression $36{{x}^{2}}+25{{y}^{2}}+60xy$ . We will get
$36{{\left( \dfrac{1}{5} \right)}^{2}}+25{{\left( \dfrac{4}{3} \right)}^{2}}+60\times \dfrac{1}{5}\times \dfrac{4}{3}$
Let us take the squares of the terms. We will get
$36\times \dfrac{1}{25}+25\times \dfrac{16}{9}+60\times \dfrac{1}{5}\times \dfrac{4}{3}$
Let us cancel 60 and 3 from the third term. We will get
$36\times \dfrac{1}{25}+25\times \dfrac{16}{9}+20\times \dfrac{1}{5}\times 4$
We can now cancel 20 and 5 from third term that results in
$36\times \dfrac{1}{25}+25\times \dfrac{16}{9}+4\times 4$
We can write this as
$\dfrac{36}{25}+25\times \dfrac{16}{9}+4\times 4$
We should multiply 25 with 16 in the second term. We will get
$\dfrac{36}{25}+\dfrac{400}{9}+4\times 4$
By solving the last term, we will get
$\dfrac{36}{25}+\dfrac{400}{9}+16$
Now, let us take the LCM of 25,9 and 1.
$$\eqalign{
  & 3\left| \!{\underline {\,
  {25,9,1} \,}} \right. \cr
  & 3\left| \!{\underline {\,
  {25,3,1} \,}} \right. \cr
  & 5\left| \!{\underline {\,
  {25,1,1} \,}} \right. \cr
  & 5\left| \!{\underline {\,
  {5,1,1} \,}} \right. \cr
  & {\text{ 1,1,1 }} \cr} $$
Hence, the LCM is $3\times 3\times 5\times 5=225$
Now, let us multiply the numerator and denominator of each term so that the denominator becomes 225.
$\dfrac{36\times 9}{25\times 9}+\dfrac{400\times 25}{9\times 25}+\dfrac{16\times 225}{1\times 225}$
Let us solve the first term to get
$\dfrac{324}{225}+\dfrac{400\times 25}{9\times 25}+\dfrac{16\times 225}{1\times 225}$
Now, we can solve the second term that results in
$\dfrac{324}{225}+\dfrac{10000}{225}+\dfrac{16\times 225}{1\times 225}$
We can multiply the third term. We will get
$\dfrac{324}{225}+\dfrac{10000}{225}+\dfrac{3600}{225}$
Now, let us combine the numerators. We will get
$\dfrac{324+10000+3600}{225}$
Now, add the numerators to get
$\dfrac{13924}{225}$
Writing this in the decimal form, we will get
$61.884$
Hence, the value of $36{{x}^{2}}+25{{y}^{2}}+60xy$ when $x=\dfrac{1}{5}\text{ and }y=\dfrac{4}{3}$ is $\dfrac{13924}{225}$ or $61.884$ .

Note: You may make errors when taking the LCM. You must take the LCM of the denominators. We have used the method of division, but you can also use the prime factorisation method to find the LCM. You must always try to cancel the common terms first to make the expression simpler. This question is solved through simple operations like squaring, multiplication, addition and division. Be careful when doing this as a small mistake can make the entire solution wrong.