plzzzz solve it......
Answers
Answer:
✮ Answer ✮
\begin{gathered}\boxed{\underline{\boxed{\bold{x = \dfrac{(a^2 + b + ab)}{(a+b)}}}}}\\\\\\\text{And,}\\\boxed{\underline{\boxed{\bold{y = -\dfrac{2ab}{(a+b)}}}}}\\\end{gathered}
x=
(a+b)
(a
2
+b+ab)
And,
y=−
(a+b)
2ab
✮Step-by-step explanation✮
\begin{gathered}\text{Given pair of linear equations in two variables:}\\\\(a-b)x + (a+b)y = a^2 - 2ab - b^2\\\\(a+b)(x+y) = a^2 + b^2\\\\\text{Simplifying the second equation, we get}\\\\(a+b)x + (a+b)y = a^2 + b^2\\\\\text{A standard pair of linear equations in two variables is}\\\\\text{of the form:}\\\\A_{1}x + B_{1}y = C_{1}\\\\\text{and}\\\\A_{2}x + B_{2}y = C_{2}\\\\\text{On comparing the given pair of linear equations}\\\\\text{in two variables with a standard pair of linear}\\\end{gathered}
Given pair of linear equations in two variables:
(a−b)x+(a+b)y=a
2
−2ab−b
2
(a+b)(x+y)=a
2
+b
2
Simplifying the second equation, we get
(a+b)x+(a+b)y=a
2
+b
2
A standard pair of linear equations in two variables is
of the form:
A
1
x+B
1
y=C
1
and
A
2
x+B
2
y=C
2
On comparing the given pair of linear equations
in two variables with a standard pair of linear
\begin{gathered}\\\text{equations in two variables, we get}\\\\\begin{array}{| c | c | c | c | c| c |}\cline{1-6} A_1 & (a-b) & B_1 & (a+b) & C_1 & (a^2-2ab-b^2) \\\cline{1-6} A_2 & (a+b) & B_2 & (a+b) & C_2 & (a^2 + b^2) \\\cline{1-6}\end{array}\\\\\\\text{Using the cross multiplication formula for solving}\\\\\text{a pair of linear equations in two variables, we get}\\\\\dfrac{x}{B_1 C_2 - B_2 C_1} = \dfrac{y}{C_1 A_2 - C_2 A_1} = -\dfrac{1}{A_1 B_2 - A_2 B_1}\\\\\\\end{gathered}
equations in two variables, we get
\cline1−6A
1
\cline1−6A
2
\cline1−6
(a−b)
(a+b)
B
1
B
2
(a+b)
(a+b)
C
1
C
2
(a
2
−2ab−b
2
)
(a
2
+b
2
)
Using the cross multiplication formula for solving
a pair of linear equations in two variables, we get
B
1
C
2
−B
2
C
1
x
=
C
1
A
2
−C
2
A
1
y
=−
A
1
B
2
−A
2
B
1
1
\begin{gathered}\implies \dfrac{x}{(a+b)(a^2 + b^2) - (a+b)(a^2 - 2ab - b^2)}\\\\\\= \dfrac{y}{(a^2 - 2ab - b^2)(a+b) - (a^2 + b^2)(a-b)}\\\\\\= - \dfrac{1}{(a-b)(a+b) - (a+b)(a+b)}\\\\\\\\\implies \dfrac{x}{a^3 + ab^2 + a^2b + b^3 - (a^3 -2a^2b - ab^2 + a^2b - 2ab^2 -b^3)}\\\\\\= \dfrac{y}{a^3 -2a^2b - ab^2 + a^2b - 2ab^2 - b^3 - (a^3 + ab^2 -a^2b - b^3)}\\\\\\= - \dfrac{1}{a^2 - b^2 - (a^2+ b^2+ 2ab)}\\\\\\\\\implies \dfrac{x}{a^3 + b^3 + a^2b + ab^2 - (a^3 - b^3 -a^2b - 3ab^2)}\\\\\\\end{gathered}
⟹
(a+b)(a
2
+b
2
)−(a+b)(a
2
−2ab−b
2
)
x
=
(a
2
−2ab−b
2
)(a+b)−(a
2
+b
2
)(a−b)
y
=−
(a−b)(a+b)−(a+b)(a+b)
1
⟹
a
3
+ab
2
+a
2
b+b
3
−(a
3
−2a
2
b−ab
2
+a
2
b−2ab
2
−b
3
)
x
=
a
3
−2a
2
b−ab
2
+a
2
b−2ab
2
−b
3
−(a
3
+ab
2
−a
2
b−b
3
)
y
=−
a
2
−b
2
−(a
2
+b
2
+2ab)
1
⟹
a
3
+b
3
+a
2
b+ab
2
−(a
3
−b
3
−a
2
b−3ab
2
)
x
\begin{gathered}= \dfrac{y}{a^3 - b^3 - a^2b - 3ab^2 - a^3 - ab^2 + a^2b + b^3}\\\\\\= - \dfrac{1}{a^2 - b^2 -a^2 - b^2 - 2ab}\\\\\\\\\implies \dfrac{x}{a^3 + b^3 + a^2b + ab^2 -a^3 + b^3 +a^2b + 3ab^2}\\\\\\= \dfrac{y}{-4ab^2} = - \dfrac{1}{-2b^2 - 2ab}\\\\\\\\\implies \dfrac{x}{2b^3 + 2a^2b +4ab^2} = -\dfrac{y}{4ab^2} = \dfrac{-1}{-2b(a+b)}\\\\\\\implies \dfrac{x}{2b(b + a^2 + ab)} = -\dfrac{y}{4ab^2} = \dfrac{1}{2b(a+b)}\\\\\\\\\end{gathered}
=
a
3
−b
3
−a
2
b−3ab
2
−a
3
−ab
2
+a
2
b+b
3
y
=−
a
2
−b
2
−a
2
−b
2
−2ab
1
⟹
a
3
+b
3
+a
2
b+ab
2
−a
3
+b
3
+a
2
b+3ab
2
x
=
−4ab
2
y
=−
−2b
2
−2ab
1
⟹
2b
3
+2a
2
b+4ab
2
x
=−
4ab
2
y
=
−2b(a+b)
−1
⟹
2b(b+a
2
+ab)
x
=−
4ab
2
y
=
2b(a+b)
1
✮ \begin{gathered}\text{On comparing $\dfrac{x}{2b(b + a^2 + ab)}$ with $\dfrac{1}{2b(a+b)}$, we get}\\\\\\x = \dfrac{2b(a^2 + b + ab)}{2b(a+b)}\\\\\\\implies \boxed{\underline{\boxed{\bold{x = \dfrac{(a^2 + b + ab)}{(a+b)}}}}}\\\\\\\\\text{And on comparing $-\dfrac{y}{4ab^2}$ with $\dfrac{1}{2b(a+b)}$, we get}\\\\\\y = \dfrac{-4ab^2}{2b(a+b)}\\\\\\\implies \boxed{\underline{\boxed{\bold{y = -\dfrac{2ab}{(a+b)}}}}}\end{gathered}
On comparing
2b(b+a
2
+ab)
x
with
2b(a+b)
1
, we get
x=
2b(a+b)
2b(a
2
+b+ab)
⟹
x=
(a+b)
(a
2
+b+ab)
And on comparing −
4ab
2
y
with
2b(a+b)
1
, we get
y=
2b(a+b)
−4ab
2
⟹
y=−
(a+b)
2ab
Given Final velocity (v) = 50 m/s
Initial velocity (u) = 30 m/s
Acceleration (a) = 2.5 m/s^2
We know v = u + at
∴ 50 = 30 + 2.5 t.
⇒ 50 - 30 = 2.5 t
⇒ 2.5/20
= t or 8 = t
∴ Time taken to accelerate = 8 seconds