{"@context":"https://schema.org/","@type":"Quiz","name":"

$\\dfrac{2}{x-2}+\\dfrac{3}{x+5}=\\dfrac{rx+t}{(x-2)(x+5)}$

The equation above is true for all $x>2$, where $r$ and $t$ are positive constants. What is the value of $rt$?

","hasPart":{"@type":"Question","typicalAgeRange":"12-22","educationalAlignment":[{"@type":"AlignmentObject","alignmentType":"educationalSubject","targetName":"SAT"},{"@type":"AlignmentObject","alignmentType":"educationalLevel","targetName":"SAT","educationalFramework":"Common Core","targetUrl":"https://classnotes.xyz/mcqs/sat-math-advanced-math-multiple-choice-question-24"}],"about":{"@type":"Thing","name":"Math, Advanced Math, Duplicate"},"educationalLevel":"intermediate","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","assesses":["Math","Advanced Math","Duplicate"],"name":"

$\\dfrac{2}{x-2}+\\dfrac{3}{x+5}=\\dfrac{rx+t}{(x-2)(x+5)}$

The equation above is true for all $x>2$, where $r$ and $t$ are positive constants. What is the value of $rt$?

","comment":{"@type":"Comment","text":"

Choice C is correct. To express the sum of the two rational expressions on the left-hand side of the equation as the single rational expression on the right-hand side of the equation, the expressions on the left-hand side must have the same denominator. Multiplying the first expression by $\\dfrac{x+5}{x-5}$ results in $\\dfrac{2(x+5)}{(x-2)(x+5)} $, and multiplying the second expression by $\\dfrac{x-2}{x-2}$ results in $\\dfrac{3(x-2)}{(x-2)(x+5)}$, so the given equation can be rewritten as

$\\dfrac{2(x+5)}{(x-2)(x+5)}+\\dfrac{3(x-2)}{(x-2)(x+5)}=\\dfrac{rx+t}{(x-2)(x+5)}$

$\\dfrac{2x+10}{(x-2)(x+5)}+\\dfrac{3x-6}{(x-2)(x+5)}=\\dfrac{rx+t}{(x-2)(x+5)}$

Since the two rational expressions on the left-hand side of the equation have the same denominator as the rational expression on the right-hand side of the equation, it follows that $(2x+10)+(3x-6)=r+t$. Combining like terms on the left-hand side yields $5x+4=rx+t$, so it follows that $r=5$ and $t=4$. Therefore, the value of $rt$ is $(5)(4)=20$.

Choice A is incorrect and may result from an error when determining the sign of either r or t. Choice B is incorrect and may result from not distributing the 2 and 3 to their respective terms in $\\dfrac{2(x+5)}{(x-2)(x+5)}+\\dfrac{3(x-2)}{(x-2)(x+5)}=\\dfrac{rx+t}{(x-2)(x+5)}$. Choice D is incorrect and may result from a calculation error.

"},"encodingFormat":"text/markdown","text":"

$\\dfrac{2}{x-2}+\\dfrac{3}{x+5}=\\dfrac{rx+t}{(x-2)(x+5)}$

The equation above is true for all $x>2$, where $r$ and $t$ are positive constants. What is the value of $rt$?

","suggestedAnswer":[{"@type":"Answer","position":0,"encodingFormat":"text/html","text":"

$-20$

"},{"@type":"Answer","position":1,"encodingFormat":"text/html","text":"

$15$

"},{"@type":"Answer","position":3,"encodingFormat":"text/html","text":"

$60$

"}],"acceptedAnswer":{"@type":"Answer","position":2,"encodingFormat":"text/html","text":"

$20$

","comment":{"@type":"Comment","text":"

$\\dfrac{2x+10}{(x-2)(x+5)}+\\dfrac{3x-6}{(x-2)(x+5)}=\\dfrac{rx+t}{(x-2)(x+5)}$

"},"answerExplanation":{"@type":"Comment","text":"

$\\dfrac{2x+10}{(x-2)(x+5)}+\\dfrac{3x-6}{(x-2)(x+5)}=\\dfrac{rx+t}{(x-2)(x+5)}$

"}}},"url":"https://classnotes.xyz/mcqs/sat-math-advanced-math-multiple-choice-question-24"}

$\\dfrac{2}{x-2}+\\dfrac{3}{x+5}=\\dfrac{rx+t}{(x-2)(x+5)}$The equation above is true for all $x>2$, where $r$ and $t$ are positive constants. What is the value of $rt$?

$\\dfrac{2}{x-2}+\\dfrac{3}{x+5}=\\dfrac{rx+t}{(x-2)(x+5)}$The equation above is true for all $x>2$, where $r$ and $t$ are positive constants. What is the value of $rt$?

$\\dfrac{2}{x-2}+\\dfrac{3}{x+5}=\\dfrac{rx+t}{(x-2)(x+5)}$The equation above is true for all $x>2$, where $r$ and $t$ are positive constants. What is the value of $rt$?

$\\dfrac{2}{x-2}+\\dfrac{3}{x+5}=\\dfrac{rx+t}{(x-2)(x+5)}$

The equation above is true for all $x>2$, where $r$ and $t$ are positive constants. What is the value of $rt$?

$\\dfrac{2}{x-2}+\\dfrac{3}{x+5}=\\dfrac{rx+t}{(x-2)(x+5)}$

The equation above is true for all $x>2$, where $r$ and $t$ are positive constants. What is the value of $rt$?

$\\dfrac{2}{x-2}+\\dfrac{3}{x+5}=\\dfrac{rx+t}{(x-2)(x+5)}$

The equation above is true for all $x>2$, where $r$ and $t$ are positive constants. What is the value of $rt$?

$\dfrac{2}{x-2}+\dfrac{3}{x+5}=\dfrac{rx+t}{(x-2)(x+5)}$The equation above is true for all $x>2$, where $r$ and $t$ are positive constants. What is the value of $rt$?

$\dfrac{2}{x-2}+\dfrac{3}{x+5}=\dfrac{rx+t}{(x-2)(x+5)}$

The equation above is true for all $x>2$, where $r$ and $t$ are positive constants. What is the value of $rt$?