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3 years ago

Solve using substitution: 2x - 3y = -1 y = x - 1

Mathematics

1 answer:

denis-greek [22]3 years ago

8 0

Answer:

x=4 y=3

Step-by-step explanation:

rewrite the equations- y=x-1 2x-3y=-1

solve y=x-1 for y

y=x-1

sub for x-1 for y in 2x-3y=-1

2x-3y=-1

2x-3(x-1)=-1

-x+3=-1

-x+3+-3=-1+-3

-x=-4

divide both sides by -4

x=4

sub 4 for x in y=x-1

y=x-1

y=4-1

y=3

WE ARE VENOM

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If you're using the app, try seeing this answer through your browser: brainly.com/question/2752942

_______________

Solve the equation:

$\mathsf{\dfrac{8}{x-5}-\dfrac{9}{x-4}=\dfrac{5}{x^2-9x+20}\qquad\qquad(x\ne 5~and~x\ne 4)}$

Reduce the fractions at the left side so that they have the same denominator:

$\mathsf{\dfrac{8(x-4)}{(x-5)(x-4)}-\dfrac{9(x-5)}{(x-4)(x-5)}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32}{x^2-4x-5x+20}-\dfrac{9x-45}{x^2-4x-5x+20}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32}{x^2-9x+20}-\dfrac{9x-45}{x^2-9x+20}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32-(9x-45)}{x^2-9x+20}=\dfrac{5}{x^2-9x+20}}$

Numerators must be equal:

$\mathsf{8x-32-(9x-45)=5}\\\\
\mathsf{8x-32-9x+45=5}\\\\
\mathsf{8x-9x=5+32-45}\\\\
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\mathsf{x=8}\quad\longleftarrow\quad\textsf{this is the solution.}$

I hope this helps. =)

Tags: rational equation fraction solution algebra

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