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Harman [31]
4 years ago
10

Solve system of equations 2x + 2y + 5z = 7 6x + 8y + 5z = 9 2x + 3y + 5z = 6

Mathematics
1 answer:
svet-max [94.6K]4 years ago
5 0

Answer:

the values of x, y and z are x= 2, y =-1 and z=1

Step-by-step explanation:

We need to solve the following system of equations.

We will use elimination method to solve these equations and find the values of x, y and z.

2x + 2y + 5z = 7     eq(1)

6x + 8y + 5z = 9     eq(2)

2x + 3y + 5z = 6     eq(3)

Subtracting eq(1) and eq(3)

2x + 2y + 5z = 7

2x + 3y + 5z = 6

-    -       -        -

_____________

0 -y + 0 = 1

-y = 1

=> y = -1

Subtracting eq(2) and eq(3)

6x + 8y + 5z = 9    

2x + 3y + 5z = 6    

-     -      -          -

______________

4x + 5y +0z = 3    

4x + 5y = 3      eq(4)

Putting value of y = -1 in equation 4

4x + 5y = 3

4x + 5(-1) = 3

4x -5 = 3

4x = 3+5

4x = 8

x= 8/4

x = 2

Putting value of x=2 and y=-1 in eq(1)

2x + 2y + 5z = 7

2(2) + 2(-1) + 5z = 7

4 -2 + 5z = 7

2 + 5z = 7

5z = 7 -2

5z = 5

z = 5/5

z = 1

So, the values of x, y and z are x= 2, y =-1 and z=1

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Or what value of x does 4x=(1/8)^x+5?<br> –15<br> –3<br> 3<br> 15
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4^x=\left(\dfrac{1}{8}\right)^{x+5}\qquad\text{use}\ a^{-1}=\dfrac{1}{a}\\\\4^x=(8^{-1})^{x+5}\\\\(2^2)^x=\bigg((2^3)^{-1}\bigg)^{x+5}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\2^{2x}=2^{-3(x+5)}\iff2x=-3(x+5)\qquad\text{use the distributive property}\\\\2x=-3x-15\qquad\text{add}\ 3x\ \text{to both sides}\\\\5x=-15\qquad\text{divide both sides by 5}\\\\x=-3

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