05eviebee
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Find $A$ and $B$ such that
\[\frac{3x+5}{x^2-x-42}=\frac{A}{x-7}+\frac{B}{x+6}.\]Write your answer in the form $(A,B)$.

Respuesta :

LammettHash

Assume

[tex]\dfrac{3x+5}{x^2-x-42}=\dfrac{3x+5}{(x-7)(x+6)}=\dfrac A{x-7}+\dfrac B{x+6}[/tex]

Combining the fractions on the right hand side gives

[tex]\dfrac{3x+5}{(x-7)(x-6)}=\dfrac{A(x-6)+B(x-7)}{(x-7)(x-6)}[/tex]

The fractions will be equal as long as the their numerator are equal:

[tex]3x+5=A(x-6)+B(x-7)[/tex]

[tex]3x+5=(A+B)x+(-6A-7B)[/tex]

Polynomials are equal to one another if the coefficients of like terms are equal:

[tex]\begin{cases}A+B=3\\-6A-7B=5\end{cases}[/tex]

Solving this, you'd get [tex]A=26[/tex] and [tex]B=-23[/tex], so that your answer would be (26, -23).

- - -

Instead of solving the system of equations above, there is a trick that involves picking [tex]x[/tex] so that some terms disappear and solving for either [tex]A,B[/tex] is much faster.

At the point where we have

[tex]3x+5=A(x-6)+B(x-7)[/tex]

notice that setting [tex]x=7[/tex] will eliminate [tex]B[/tex]. Doing so, we get

[tex]3(7)+5=A(7-6)\implies26=A[/tex]

while setting [tex]x=6[/tex] would give

[tex]3(6)+5=B(6-7)\implies23=-B\implies B=-23[/tex]

StarGazers

Answer:

We factor the denominator in the left-hand side to get

[tex]\[\frac{3x+5}{(x-7)(x+6)}= \frac{A}{x - 7} + \frac{B}{x + 6}.\][/tex]

We then multiply both sides by

[tex]$(x - 7)(x + 6)$[/tex], to get [tex]\[3x + 5 = A(x + 6) + B(x - 7).\][/tex]

We can solve for [tex]$A$[/tex] and [tex]$B$[/tex] by substituting suitable values of [tex]$x$[/tex].

For example, setting [tex]$x = 7$[/tex], the equation becomes [tex]$26 = 13A$[/tex],

so [tex]$A = 2$[/tex].

Setting [tex]$x = -6$[/tex], the equation becomes [tex]$-13 = -13B$[/tex],

so [tex]$B = 1$[/tex]

Therefore, [tex]$(A,B) = \boxed{(2,1)}$[/tex].

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