
If the value of the equation . Find the value of .
(a) 12
(b) 25
(c) 33
(d) 47
(e) 86
Answer
511.8k+ views
Hint: Here we have an equation , using this equation we first find the value of ‘x’. To find the value of ‘x’, we need to add every term given in the equation which are coefficients of x. After adding we divide the both sides with the value that is equal to the coefficient of ‘x’ to get the value of x. Once, we get the value of ‘x’, we substitute in to get the required answer.
Complete step-by-step answer:
Given that the value of the equation . We need to find the value of .
Let us first find the value of ‘x’ using the equation .
So, we have .
Adding the all terms which are coefficients of ‘x’, we get
.
.
Dividing with 19 on both sides, we get
.
We got the value of ‘x’ to be ‘6’.
Now we use the obtained value of ‘x’ to find the value of . Let us represent the value of as ‘y’.
Let us substitute the value of ‘x’ in . On substituting, we get
.
.
.
We got the value of as 33.
∴ The value of ‘x’ is 6 and the value of is 33.
So, the correct answer is “Option c”.
Note: We should not make calculation mistakes while finding the value of ‘x’. We need to make division with exact precision to get the value of ‘x’. Similar types of problems will be given by changing the degree of ‘x’ in either of the equations, adding an extra constant to the equation. Similar procedures will be followed even to solve such types of problems.
Complete step-by-step answer:
Given that the value of the equation
Let us first find the value of ‘x’ using the equation
So, we have
Adding the all terms which are coefficients of ‘x’, we get
Dividing with 19 on both sides, we get
We got the value of ‘x’ to be ‘6’.
Now we use the obtained value of ‘x’ to find the value of
Let us substitute the value of ‘x’ in
We got the value of
∴ The value of ‘x’ is 6 and the value of
So, the correct answer is “Option c”.
Note: We should not make calculation mistakes while finding the value of ‘x’. We need to make division with exact precision to get the value of ‘x’. Similar types of problems will be given by changing the degree of ‘x’ in either of the equations, adding an extra constant to the equation. Similar procedures will be followed even to solve such types of problems.
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