Question

Solve by elimination method 5x-2y=11, 3x+4y=4.​
Need Help;)​

Answer 1

Answer:

➙5x-2y=11

➙3x+4y=4

We have given two Equation we have to make one Variable of both equation same so that it can be cancelled.

➙5x-2y=11----×3

➙3x+4y=4----×5

New Equation obtained

➙15x-6y=33

➙15x+20y=20

➙

____________

-26y=13

y= -13/26

y= -1/2

Now put y's value into this equation

➙5x-2y=11

➙5x-2×-1/2=11

➙5x+1=11

➙5x=10

➙x= 2

Check:

➙5x-2y=11

Taking LHS

➙5(2)-2×-1/2

➙10+1=11

Since,LHS = RHS(Verified)

Step-by-step explanation:

Answer 2

Step-by-step explanation:

★ How to do :-

• Here, we are given with two equations. We are asked to find the value of x and y using substitution method. By using the first equation we will find the value of x by substituting the values. Then, we use the hint of x and then we can find the value of y. Then we can find the original value of x by using the value of y. Here, we also shift numbers from one hand side to the other which changes it's sign. So, let's solve!!

➤ Solution :-

${\sf \leadsto 5x - 2y = 11 \: --- (i)}$

${\sf \leadsto 3x + 4y = 4 \: --- (ii)}$

• First let's find the value of x by using first equation.

${\sf \leadsto 5x - 2y = 11}$

• Shift the number 2y from LHS to RHS, changing it's sign.

${\sf \leadsto 5x = 11 - 2y}$

• Shift the number 5 from LHS to RHS.

${\sf \leadsto x = \dfrac{11 - 2y}{5}}$

• Now, let's find the value of y using the second equation.

Value of y :-

${\rm3x + 4y = 4}$

• Substitute the value of x.

${\sf 3 \bigg( \dfrac{11 - 2y}{5} \bigg) + 4y = 4}$

• Multiply the number 3 with both numbers in brackets.

${\sf\dfrac{33 - 6y}{5} + 4y = 4}$

• Convert the number 4y to like fraction and add it with the fraction.

${\sf\leadsto \dfrac{33 - 6y + 20y}{5} = 4}$

• Add the variable values on denominator.

${\</strong><strong>s</strong><strong>f</strong><strong> \dfrac{33 + 14y}{5} = 4}$

• Shift the number 5 from LHS to RHS.

${\sf33 + 14y = 5 \times 4}$

• Multiply the values on RHS.

${\sf \leadsto 33 + 14y = 20}$

• Shift the number 33 from LHS to RHS, changing it's sign.

${\sf \leadsto 14y = 20 - 33}$

• Subtract the values on RHS.

${\tt \leadsto 14y = (-13)}$

• Shift the number 14 from LHS to RHS.

${\sf y = \dfrac{(-13)}{14}}$

• Now, let's find the value of x by second equation.

Value of x :-

${\sf 3x + 4y = 4}$

• Substitute the value of y.

${\sf3x + 4 \bigg( \dfrac{(-13)}{14} \bigg) = 4}$

• Multiply the number 4 with the fraction in bracket.

${\sf 3x + \dfrac{(-52)}{14} = 4}$

• Shift the fraction on LHS to RHS, changing it's sign.

${\sf3x = 4 - \dfrac{(-52)}{14}}$

• Convert the values on LHS to like fractions.

${\sf 3x = \dfrac{56}{14} - \dfrac{(-52)}{14}}$

• Subtract those fractions now.

${\sf 3x = \dfrac{108}{14}}$

• Shift the number 3 from LHS to RHS.

${\sf x = \dfrac{108}{14} \div \dfrac{3}{1}}$

• Take the reciprocal of second fraction and multiply both fractions.

$\\ {\sf x = \dfrac{108}{14} \times \dfrac{1}{3}}$

• Write those fractions in lowest form by cancellation method.

${\sf x = \dfrac{\cancel{108}}{14} \times \dfrac{1}{\cancel{3}} = \dfrac{36 \times 1}{14 \times 1}}$

• Write the fraction in lowest form to get the answer.

${\sf \cancel \dfrac{36}{14} = \dfrac{18}{7}}$

${\red{\underline{\boxed{\footnotesize \sf{So, \: the \: value \: of \: x \: and \: y \: is \dfrac{18}{7} \: and \: \dfrac{(-13)}{14} \: respectively.}}}}}$