Mathematics
Easy

Question

Find the following expression with the values of x = 1 and y = 22x2 + 3y2

13    14    15    122

Hint:

The given expression is 2x2+3y2There are two variables and two constants in the above expression. The variables are: ‘x’ and ‘y’. Values assigned to the variables are x = 1 and y = 2. We have to substitute these values and find the final value of the expression.There are two terms in the above expression. The first term has variable x raised to 2. It means we have to multiply x twice. And then we have to multiply the value by a constant 2. In the second term, the variable y is raised to 2. It means we have to multiply y twice. Then we have to multiply it by constant 3.We have to first substitute the value and perform the required operations.To substitute the value, we have to replace x by 1 and y by 2. After substituting the value, we will get the following equation:2x2+3y2 = 2(1)2+ 3(2)22x2+3y2 = 2(1)(1)+3(2)(2)2x2+3y2 = 2+122x2+3y2 = 14 Therefore, the value of equation is 14So option (b), which is ‘14', is the right option.

We have to be careful about the operations because the variables. We have to follow proper order for performing the operations. The variables take different values under different conditions. So we have to careful while substituting the values. The main part of such type of questions is just replacing the variables by their value and performing the operation.