Math, asked by mangilalgurjar1947, 1 month ago

2x- 3y = -2 ,4x+y = 24 by cross multiplication

Answers

Answered by meherumakanta541
0

Step-by-step explanation:

2x-3y=-2

2x=-2+3y

x=3y-2/2

4x+y=24

4(3y-2/2)+y=24

6y-4+y=24

7y=28

y=28/7=4

x= 3y-2/2

x=3×4-2/2

x=10/2

x=5

Answered by devyanibehera5b
0

Answer:

x=

47

5

and y=−

47

23

Step-by-step explanation:

\boxed{\huge{\bf{\pink{Answer}}}}

Answer

x=\dfrac{5}{47} \ and \ y=-\dfrac{23}{47}x=

47

5

and y=−

47

23

Step-by-step explanation:

\begin{gathered}Given \ 5x-3y=2 \ and \ 4x+7y=-3\\\\\end{gathered}

Given 5x−3y=2 and 4x+7y=−3

We have to find the value of x and y by using cross multiplication method.

\begin{gathered}For \ cross \ multiplication \ method \ we \ have\\\\\dfrac{x}{b_1c_2-b_2c_1}=\dfrac{y}{c_1a_2-c_2a_1}=\dfrac{1}{a_1b_2-a_2b_1}\\\\\\5x-3y-2=0 \ and \ 4x+7y+3=0\\\\\\we \ have\\\\a_1=5 \ b_1=-3 \ and \ c_1=-2\\\\\\a_2=4 \ b_2=7 \ and \ c_2=3\\\\\\putting \ value \ in \ formula\end{gathered}

For cross multiplication method we have

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

5x−3y−2=0 and 4x+7y+3=0

we have

a

1

=5 b

1

=−3 and c

1

=−2

a

2

=4 b

2

=7 and c

2

=3

putting value in formula

\begin{gathered}\dfrac{x}{(-3\times3)-(7\times-2)}=\dfrac{y}{(-2\times4)-(3\times5)}=\dfrac{1}{(5\times7)-(4\times-3)}\\\\\\ \dfrac{x}{-9+14}= \dfrac{1}{35+12} \ and \ \dfrac{y}{-8-15}= \dfrac{1}{35+12}\\\\\\x=\dfrac{5}{47} \ and \ y=-\dfrac{23}{47}\end{gathered}

(−3×3)−(7×−2)

x

=

(−2×4)−(3×5)

y

=

(5×7)−(4×−3)

1

−9+14

x

=

35+12

1

and

−8−15

y

=

35+12

1

x=

47

5

and y=−

47

23

So we get our answer

x=\dfrac{5}{47} \ and \ y=-\dfrac{23}{47}x=

47

5

and y=−

47

23

\begin{gathered}Verification\\\\\\putting \ x=\dfrac{5}{47} \ and \ y=-\dfrac{23}{47} \ in \ 5x-3y=2\\\\\\\L.H.S.=(5\times\dfrac{5}{47})-3\times(-\dfrac{23}{47})\\\\\\L.H.S.=\dfrac{25+69}{47}=2\\\\\\L.H.S.=R.H.S\\\\Now \ putting \ in \ 4x+7y=-3\\\\\\L.H.S.=(4\times\dfrac{5}{47})+(7\times-\dfrac{23}{47})\\\\\\L.H.S.=\dfrac{20-161}{47}\\\\\\L.H.S=\dfrac{-141}{47}=-3\\\\\\L.H.S.=R.H.S\\\\\\Hence \ we \ verified \ it.\end{gathered}

Verification

putting x=

47

5

and y=−

47

23

in 5x−3y=2

\L.H.S.=(5×

47

5

)−3×(−

47

23

)

L.H.S.=

47

25+69

=2

L.H.S.=R.H.S

Now putting in 4x+7y=−3

L.H.S.=(4×

47

5

)+(7×−

47

23

)

L.H.S.=

47

20−161

L.H.S=

47

−141

=−3

L.H.S.=R.H.S

Hence we verified it.

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