Question

Solve the pair of linear equations. 2/+ = 3/2 and /2− = −3/10

x+y≠0, 2x − y≠0

pls give the correct answer

Answer 1

Answer:

blablablaieihrkdodjd

Step-by-step explanation:

b★ Initial velocity(u)=0 m/s

(As it starts from rest.)

\bigstar\ \sf{\color{teal} {Final\ velocity(v) = 3\ m/s}}★ Final velocity(v)=3 m/s

\bigstar\ \sf{\color{teal} {Time(t)=6\ s}}★ Time(t)=6 s

TO FIND :-

\sf{\color{gold} {The\ distance\ travelled.}}The distance travelled.

SOLUTION :-

\sf{\color{blue} {We\ know,}}We know,

\boxed{\sf{\color{red} {v=u+at}}}

v=u+at

\sf{\color{blue} {Putting\ the\ known\ values\ :}}Putting the known values :

\sf{\color{red} {\implies 3=0+a\times 6}}⟹3=0+a×6

\sf{\color{red} {\implies 3=6a}}⟹3=6a

\sf{\color{red} {\implies a=\frac{3}{6} }}⟹a=

6

3

\sf{\color{red} {\implies a=0.5\ m/s^2}}⟹a=0.5 m/s

2

\sf{\color{blue} {We\ also\ know\ that,}}We also know that,

\boxed{\sf{\color{red} {v^2-u^2=2as}}}

v

2

−u

2

=2as

\sf{\color{blue} {Putting\ the\ known\ values\ :}}Putting the known values :

\sf{\color{red} {\implies (3)^2-(0)^2=2\times0.5\times s}}⟹(3)

2

−(0)

2

=2×0.5×s

\begin{lgathered}\sf{\color{red} {\implies 9=1\times s}}\\\\\underline{\color{green}{\boxed{\boxed{\sf{\color{red} {\implies s=9\ m}}}}}}\end{lgathered}

⟹9=1×s

⟹s=9 m

\diamond\ \large\underline{\color{blue} {\sf{So,\ the\ distance\ travelled\ is\ 9\ m.}}}⋄

So, the distance travelled is 9 m.

Answer 2

Answer:

2xy/x+y=3/2

4xy=3x+3y-----------eq.(1)

again. xy/2x-y=-6x+3y -----eq.(2)

eq.(1) - eq.(2)

4x-10y=3x+6x+3y-3y

-6xy=9x

y=-3/2

putting value of y in eq.1

4x×-3/2=3x+3×-3/2

-6x=3x-9/2

9x=9/2

x=1/2

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