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If $14y^2-21x=a(2y^2-3x)$ , where $a$ is a constant, what is the value of $a$ ?

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Determine the direction of a vector scalar multiple

Full solution

Q. If

14y^2-21x=a(2y^2-3x)

, where

a

is a constant, what is the value of

a

?

Compare coefficients: Compare the coefficients of the like terms on both sides of the equation. $\newline$ The equation given is $14y^2 - 21x = a(2y^2 - 3x)$ . To find the value of ' $a$ ', we need to compare the coefficients of the corresponding terms on both sides of the equation.
Compare $y^2$ : Compare the coefficients of $y^2$ . On the left side, the coefficient of $y^2$ is $14$ . On the right side, the coefficient of $y^2$ is $2a$ . Therefore, we have $14 = 2a$ .
Solve for 'a': Solve for 'a' using the $y^2$ terms. $\newline$ Divide both sides of the equation $14 = 2a$ by $2$ to solve for 'a'. $\newline$ $14 \div 2 = 2a \div 2$ $\newline$ $7 = a$
Verify with x terms: Verify the solution with the x terms. $\newline$ Now, let's check if the value of 'a' is consistent with the coefficients of the x terms. On the left side, the coefficient of x is $-21$ . On the right side, the coefficient of x is $-3a$ . Substitute the value of 'a' we found into $-3a$ to see if it equals $-21$ . $\newline$ $-3 \times 7 = -21$ $\newline$ $-21 = -21$ $\newline$ This confirms that the value of 'a' is correct.

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