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exis [7]
4 years ago
8

Solve for the following system of equations. -7x+6y=9 -2x-5y+16 x=? y=?

Mathematics
2 answers:
CaHeK987 [17]4 years ago
4 0

Answer:

x=-3\\\\y=-2

Step-by-step explanation:

Given the system of equations \left \{ {{-7x+6y=9} \atop {-2x-5y=16}} \right., you can use the Elimination Method to solve it.

You can multiply the first equation by -2 and the second equation by 7, then add both equations and solve for the variable "y":

\left \{ {{14x-12y=-18} \atop {-14x-35y=112}} \right.\\...........................\\-47y=94\\\\y=\frac{94}{-47}\\\\y=-2

Substitute the value of "y" into any original equations and solve for the variable "x". Then:

-7x+6y=9\\\\-7x+6(-2)=9\\\\-7x-12=9\\\\-7x=9+12\\\\x=\frac{21}{-7}\\\\x=-3

Alex17521 [72]4 years ago
3 0

Answer:

x = -3 and y = -2

Step-by-step explanation:

It is given that,

-7x + 6y = 9     ----(1)

-2x - 5y = 16    -------(2)

<u>To find the solution of given equations</u>

eq(1) * 2 ⇒

-14x + 12y = 18 ------(3)

eq(2) * 7 ⇒

-14x - 35y = 112   ---(4)

eq (3) - eq(4) ⇒

-14x + 12y = 18 ------(3)

<u>-14x - 35y = 112 </u>  ---(4)

    0   4y = -94

y = 94/(-47) = -2

Substitute the value of y in eq (2)

-2x - 5y = 16    -------(2)

-2x - 5*-2 = 16

-2x +10 = 16

-2x = 6

x = 6/-2 = -3

Therefore x = -3 and y = -2

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What are the roots of the polynomial equation x^3-5x+5=2x^2-5? Use a graphing calculator and a system of equations. Round nonint
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The right answer is c. –2.24, 2, 2.24


This question needs to be solved in two ways. First, using a graphing calculator. Next, using a system of equations.


1. Using a graphing calculator.


We have the following polynomial equation:

x^3-5x+5=2x^2-5


By ordering this equation we have:

x^3-2x^2-5x+10=0


So, we can say that this equation comes from a function given by:

f(x)=x^3-2x^2-5x+10


Thus, by plotting this function, we have that the graph of this function is indicated in Figure 1. By zooming, we can see, in Figure 2, that the roots of the polynomial equation are the x-intercepts of f(x) which are:


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Finally, rounding noninteger roots to the nearest hundredth we have:


\boxed{Root_{1}=-2.24} \\ \\ \boxed{Root_{2}=2} \\ \\ \boxed{Root_{3}=2.24}


2. Using a system of equations.


The ordered equation is:

x^3-2x^2-5x+10=0


By arranging to factor out we have:

x^3-5x-2x^2+10=0


Then, by factoring:

x(x^2-5)-2(x^2-5)=0


Term (x^2-5) is a common factor, thus:


(x-2)(x^2-5)=0 \\ \\ (x-2)(x-\sqrt{5})(x+\sqrt{5})=0 \\ \\ Finally: \\ \\ \boxed{Root_{1}=-\sqrt{5}=-2.24} \\ \\ \boxed{Root_{2}=2} \\ \\ \boxed{Root_{3}=\sqrt{5}=2.24}

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