Question

solve by elimination method 2x+3y-11=0,2x-4y+24=0​

Answer 1

Step-by-step explanation:

with use of elimination method , we get

(x,y)=(-2,5)

Answer 2

$\large\sf\underline{Given\::}$

Equation 1 :

• 2x + 3y - 11 = 0

Equation 2 :

• 2x - 4y + 24 = 0

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$\large\sf\underline{To~find\::}$

• Value of x and y.

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$\large\sf\underline{Concept\::}$

Simultaneous equation can be solved by three methods those are elimination method , comparison method , and substitution method . So we need to go for elimination method .

In this [ elimination ] method we need to first make sure that the leading coefficient of two equations are same. If it is not same we need to make it same by multiplying with suitable numbers. Secondly we need to subtract equation (ii) from equation (i) . Doing so we would get the value of y and then substituting the value of y in any of the two equation we would get the value of x. I hope am clear now , Let's begin !

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$\large\sf\underline{Solution\::}$

$\sf\:2x + 3y - 11 = 0$

• Transposing -11 to RHS it becomes +11

$\sf\implies\:2x + 3y = 11$ --- ( Equation i )

$\sf\:2x - 4y + 24 = 0$

• Transposing +24 to RHS it becomes -24

$\sf\implies\:2x - 4y = -24$ --- ( Equation ii )

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Now we are observing that the leading coefficient is same in both the equations so we directly jump to the second step where we need to subtract the equation (ii) from equation (i) .

$\sf\:equation~(i)~-~equation~(ii)~$

Subtraction of equation should be done as :

• LHS - LHS = RHS - RHS

$\sf\implies\:2x+3y-(2x-4y)=11-(-24)$

• Multiplying and removing the brackets

$\sf\implies\:2x+3y-2x+4y=11+24$

• Arranging the terms for proceeding with simple calculations

$\sf\implies\:2x-2x+3y+4y=11+24$

$\sf\implies\:\cancel{2x}-\cancel{2x}+3y+4y=11+24$

$\sf\implies\:3y+4y=11+24$

$\sf\implies\:7y=11+24$

$\sf\implies\:7y=35$

• Transposing 7 to RHS it goes to the denominator

$\sf\implies\:y=\frac{35}{7}$

• Reducing the fraction to the lower terms

$\sf\implies\:y=\cancel{\frac{35}{7}}$

$\small{\underline{\boxed{\mathrm\red{\implies\:y\:=\:5}}}}$ ★

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Now substituting the value of y as 5 in the equation (i) :

Equation (i) : $\sf\:2x + 3y = 11$

$\sf\implies\:2x + 3(5) = 11$

$\sf\implies\:2x + 15 = 11$

• Transposing +15 to RHS it becomes -15

$\sf\implies\:2x= 11-15$

$\sf\implies\:2x= -4$

• Transposing 2 to RHS it goes to the denominator

$\sf\implies\:x=\frac{-4}{2}$

• Reducing the fraction to the lower terms

$\sf\implies\:x=\cancel{\frac{-4}{2}}$

$\small{\underline{\boxed{\mathrm\red{\implies\:x\:=\:(-2)}}}}$ ★

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So we got :

• x = ( - 2 ) and

• y = 5

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Verifying :

Now let's check if our answers are correct. In order to do so we would substitute both the values of x and y in any of the two equation. Doing so if we get LHS = RHS, our answers would be correct.

Taking [equation i] :

$\sf\:2x + 3y = 11$

• Substituting the values of x and y

$\sf\to\:2(-2) + 3(5) = 11$

• Multiplying and removing the brackets

$\sf\to\:-4 + 15= 11$

$\sf\to\:11= 11$

$\bf\to\:LHS= RHS$

$\small\fbox\purple{Hence~Verified~!! }$

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Required anSwers :

• Value of x $\green{\mid{\fbox{\tt{(~-2~)}}\mid}}$‎

• Value of y $\green{\mid{\fbox{\tt{5}}\mid}}$‎

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!! Hope it helps !!