Question Video: Finding the Value of a Constant Using Division by a Monomial Mathematics • 10th Grade

Find the value of the constant π‘˜, given that (4π‘₯⁡𝑦² + π‘˜π‘₯⁷𝑦³)/2π‘₯²𝑦 = 2π‘₯³𝑦 + 3π‘₯⁡𝑦².

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Video Transcript

Find the value of the constant π‘˜, given that four π‘₯ raised to the fifth power 𝑦 squared plus π‘˜π‘₯ raised to the seventh power 𝑦 cubed all over two π‘₯ squared 𝑦 is equal to two π‘₯ cubed 𝑦 plus three π‘₯ raised to the fifth power 𝑦 squared.

In this question, we are asked to find the value of a constant π‘˜ given an equation involving this unknown constant. To answer this question, it is worth recalling that when we say β€œequals” in this case, we are really asking for the expressions to be equivalent. In other words, we want both sides of the equation to be equal for any valid values of π‘₯ and 𝑦.

We can find the value of π‘˜ by evaluating the division on the left-hand side of the equation. To do this, we will start by dividing each term in the numerator separately by the denominator. We can then simplify by recalling that the quotient rule for exponents tells us if a base 𝑏 is nonzero, then 𝑏 raised to the power of π‘š over 𝑏 raised to the power of 𝑛 is equal to 𝑏 raised to the power of π‘š minus 𝑛.

We can use this result to simplify each division, where we note that π‘₯ and 𝑦 are nonzero since we cannot divide by zero. We can rewrite 𝑦 as 𝑦 raised to the first power and apply the quotient rule for exponents on the first term to obtain two π‘₯ raised to the power of five minus two 𝑦 raised to the power of two minus one.

We can apply the same process to the second term to get π‘˜ over two times π‘₯ raised to the power of seven minus two 𝑦 raised to the power of three minus one. We can then evaluate the expressions in the exponents to obtain the following equation. We can then note that the first terms on both sides of the equation are identical.

Next, we can see that the second terms have the same variables raised to the same exponents. Therefore, for the expressions to be equivalent, the coefficients of these terms must be equal. Hence, π‘˜ over two must be equal to three, which we can solve to show that π‘˜ equals six.

It is worth noting that since we want both sides of the equation to be identical, we can verify this answer by substituting valid values of π‘₯ and 𝑦 into both sides of the equation. For instance, let’s substitute π‘₯ equals one and 𝑦 equals one. If we did this, then all of the variables become factors of powers of one. So the equation simplifies to give four plus π‘˜ all over two is equal to two plus three. We can then solve this equation for π‘˜, and we find that π‘˜ equals six.

It is worth noting that this is not a proof that π‘˜ is equal to six, since we need to show it is true for all valid values of π‘₯ and 𝑦. And this method only considers one such pair of values. However, it does help verify that our answer of π‘˜ equals six is correct.

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