Question

6. If a - b = 3 and a + b = 5, find
(b) a2 + b2
(a) ab​

Answer 1

$\huge\bf{\red{\underline{\underline{\mathcal{AnSwer}}}}}$

Given:-

• a-b = 3 ... eq.1
• a+b = 5 ... eq.2

To find:-

• a²+b²
• ab

Solution:-

adding eq. 1 and 2 to find the values of a and b

$\implies$ $\sf a-b+a+b=3+5$

$\implies$ $\sf 2a=8$

$\implies$ $\sf a = 4$

putting the value of a in eq. 1

$\implies$ $\sf 4+b=3$

$\implies$ $\sf -b = -1$

$\implies$ $\sf b = 1$

now, let's find the value of a²+b²

$\implies$ a²+b² = (4)²+(1)²

$\implies$ 4+1 = 5

now, let's find the value of ab

$\implies$ a×b = 4×1 = 4

Answer 2

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{a + b = 5} \\ \\ &\sf{a - b = 3} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  Find :-  \begin{cases} &\sf{ {a}^{2} + {b}^{2} } \\ \\ &\sf{ab}  \end{cases}\end{gathered}\end{gathered}$

$\large\underline\purple{\bold{Solution :- }}$

Given

$:  \implies \tt \: a + b = 5 \: - - (i)$

$:  \implies \tt \: a - b = 3 \: - - (ii)$

On adding (i) and (ii) equations, we get

$:  \implies \tt \: a + b + a - b = 5 + 3$

$:  \implies \tt \: 2a = 8$

$:  \implies \tt \: a = \dfrac{ \cancel8}{ \cancel2} = 4$

$:  \implies \boxed{ \red{ \tt \: a \: = \: 4}}$

On substituting a = 4, in equation (i), we get

$:  \implies \tt \: 4 + b = 5$

$:  \implies \tt \: b = 5 - 4$

$:  \implies \boxed{ \red{\tt \: b \: = \: 1}}$

Now,

$\red{ \bf \: To \: find : \: {a}^{2} + {b}^{2} }$

On substituting the values of a and b, we get

$:  \implies \tt \: {4}^{2} + {1}^{2}$

$:  \implies \tt \: 16 + 1$

$:  \implies \tt \: 17$

$:  \implies \tt \: \boxed{ \purple{ \bf \: {a}^{2} + {b}^{2} = 17}}$

$\red{ \bf \: To \: find : - \: ab}$

On substituting the values of a and b, we get

$:  \implies \tt \: 4 \times 1$

$:  \implies \tt \: 4$

$:  \implies \tt \: \boxed{ \purple{ \bf \: ab \: = \: 4}}$