Solve each system by substitution. -5x + 3y = 3 and -4x + y = 1

1 answer:

5 0

$\left\{\begin{array}{ccc}-5x+3y=3\\-4x+y=1&|\text{add 4x to both sides}\end{array}\right\\\\\left\{\begin{array}{ccc}-5x+3y=3\\y=4x+1&(*)\end{array}\right\\\\\text{Substitute }\ (*)\ \text{to the first equation}\\\\-5x+3(4x+1)=3\qquad\text{use distributive property}\\\\-5x+(3)(4x)+(3)(1)=3\\\\-5x+12x+3=3\qquad\text{subtract 3 from both sides}\\\\7x=0\qquad\text{divide both sides by (-5)}\\\\\boxed{x=0}\\\\\text{Substitute the value of x to}\ (*):\\\\y=4(0)+1\\y=0+1\\\boxed{y=1}\\\\Answer:\ x=0\ and\ y=1.$

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Evaluate the definite integral. <br> 1 x4(1 + 2x5)5 dx.

Nata [24]

Answer:

Step-by-step explanation:

Given the definite integral $\int\limits {\dfrac{x^4}{(1-2x^5)^5} } \, dx$ , we to evaluate it. Using integration by substitution method.

Let u = 1-2x⁵ ...1

du/dx = -10x⁴

dx = du/-10x⁴.... 2

Substitute equation 1 and 2 into the integral function and evaluate the resulting integral as shown;

$= \int\limits {\dfrac{x^4}{u^5} } \, \dfrac{du}{-10x^4}$

$= \dfrac{-1}{10} \int\limits {\dfrac{du}{u^5} } \\\\= \dfrac{-1}{10} \int\limits {{u^{-5}du } \\= \dfrac{-1}{10} [{\frac{u^{-5+1}}{-5+1}] \\\\= \dfrac{-1}{10} ({\frac{u^{-4}}{-4})\\\\$