Respuesta :
Answer:
x = 2 y= -1/2
Step-by-step explanation:
5x-2y=11 {1} . 2
3x+4y=4 {2} . 1
-> 10x-4y=22
+ 3x+4y = 4
-> 13x = 26
-> x= 26/13 x= 2/1 x= 2
-> 3(2)+4y=4
->6+4y=4 -> 4y= 4-6
-> 4y= -2 -> y= -2/4 -> y=-1/2
★ 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!!
[tex]\:[/tex]
➤ Solution :-
[tex]{\sf \leadsto 5x - 2y = 11 \: --- (i)}[/tex]
[tex]{\sf \leadsto 3x + 4y = 4 \: --- (ii)}[/tex]
First let's find the value of x by using first equation.
[tex]{\tt \leadsto 5x - 2y = 11}[/tex]
Shift the number 2y from LHS to RHS, changing it's sign.
[tex]{\tt \leadsto 5x = 11 - 2y}[/tex]
Shift the number 5 from LHS to RHS.
[tex]{\tt \leadsto x = \dfrac{11 - 2y}{5}}[/tex]
[tex]\:[/tex]
Now, let's find the value of y using the second equation.
Value of y :-
[tex]{\tt \leadsto 3x + 4y = 4}[/tex]
Substitute the value of x.
[tex]{\tt \leadsto 3 \bigg( \dfrac{11 - 2y}{5} \bigg) + 4y = 4}[/tex]
Multiply the number 3 with both numbers in brackets.
[tex]{\tt \leadsto \dfrac{33 - 6y}{5} + 4y = 4}[/tex]
Convert the number 4y to like fraction and add it with the fraction.
[tex]{\tt \leadsto \dfrac{33 - 6y + 20y}{5} = 4}[/tex]
Add the variable values on denominator.
[tex]{\tt \leadsto \dfrac{33 + 14y}{5} = 4}[/tex]
Shift the number 5 from LHS to RHS.
[tex]{\tt \leadsto 33 + 14y = 5 \times 4}[/tex]
Multiply the values on RHS.
[tex]{\tt \leadsto 33 + 14y = 20}[/tex]
Shift the number 33 from LHS to RHS, changing it's sign.
[tex]{\tt \leadsto 14y = 20 - 33}[/tex]
Subtract the values on RHS.
[tex]{\tt \leadsto 14y = (-13)}[/tex]
Shift the number 14 from LHS to RHS.
[tex]{\tt \leadsto y = \dfrac{(-13)}{14}}[/tex]
[tex]\:[/tex]
Now, let's find the value of x by second equation.
Value of x :-
[tex]{\tt \leadsto 3x + 4y = 4}[/tex]
Substitute the value of y.
[tex]{\tt \leadsto 3x + 4 \bigg( \dfrac{(-13)}{14} \bigg) = 4}[/tex]
Multiply the number 4 with the fraction in bracket.
[tex]{\tt \leadsto 3x + \dfrac{(-52)}{14} = 4}[/tex]
Shift the fraction on LHS to RHS, changing it's sign.
[tex]{\tt \leadsto 3x = 4 - \dfrac{(-52)}{14}}[/tex]
Convert the values on LHS to like fractions.
[tex]{\tt \leadsto 3x = \dfrac{56}{14} - \dfrac{(-52)}{14}}[/tex]
Subtract those fractions now.
[tex]{\tt \leadsto 3x = \dfrac{108}{14}}[/tex]
Shift the number 3 from LHS to RHS.
[tex]{\tt \leadsto x = \dfrac{108}{14} \div \dfrac{3}{1}}[/tex]
Take the reciprocal of second fraction and multiply both fractions.
[tex]{\tt \leadsto x = \dfrac{108}{14} \times \dfrac{1}{3}}[/tex]
Write those fractions in lowest form by cancellation method.
[tex]{\tt \leadsto x = \dfrac{\cancel{108}}{14} \times \dfrac{1}{\cancel{3}} = \dfrac{36 \times 1}{14 \times 1}}[/tex]
Write the fraction in lowest form to get the answer.
[tex]{\tt \leadsto x = \cancel \dfrac{36}{14} = \dfrac{18}{7}}[/tex]
[tex]\:[/tex]
[tex]{\red{\underline{\boxed{\bf So, \: the \: value \: of \: x \: and \: y \: is \dfrac{18}{7} \: and \: \dfrac{(-13)}{14} \: respectively.}}}}[/tex]