Respuesta :

zrh2sfo

Answer:

x = 5 , y = -1

Step-by-step explanation:

Solve the following system:

{4 y + x = 1 | (equation 1)

5 y + 3 x = 10 | (equation 2)

Swap equation 1 with equation 2:

{3 x + 5 y = 10 | (equation 1)

x + 4 y = 1 | (equation 2)

Subtract 1/3 × (equation 1) from equation 2:

{3 x + 5 y = 10 | (equation 1)

0 x+(7 y)/3 = -7/3 | (equation 2)

Multiply equation 2 by 3/7:

{3 x + 5 y = 10 | (equation 1)

0 x+y = -1 | (equation 2)

Subtract 5 × (equation 2) from equation 1:

{3 x+0 y = 15 | (equation 1)

0 x+y = -1 | (equation 2)

Divide equation 1 by 3:

{x+0 y = 5 | (equation 1)

0 x+y = -1 | (equation 2)

Collect results:

Answer: {x = 5 , y = -1

Riniyah2100998

Answer:\mathrm{Range\:of\:}\frac{10-x}{52x+5}:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)<-\frac{1}{52}\quad \mathrm{or}\quad \:f\left(x\right)>-\frac{1}{52}\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:-\frac{1}{52}\right)\cup \left(-\frac{1}{52},\:\infty \:\right)\end{bmatrix}

Step-by-step explanation:\mathrm{Domain\:of\:}\:\frac{10-x}{52x+5}\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x<-\frac{5}{52}\quad \mathrm{or}\quad \:x>-\frac{5}{52}\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:-\frac{5}{52}\right)\cup \left(-\frac{5}{52},\:\infty \:\right)\end{bmatrix}

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