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Vlad1618 [11]
2 years ago
13

Solve the system of equations. 5x y = 9 3x 2y = 4 (−2, 5) (1, 4) (2, −1) (4, −4)

Mathematics
1 answer:
kobusy [5.1K]2 years ago
6 0

The solution of the system of equations that is given is (2,-1).

Given a system of equations are 5x+y=9 and 3x+2y=4.

A system of linear equations (or linear system) is a collection of one or more linear equations involving the same variables.  A solution to a linear system is an assignment of values to the variables such that all the equations are simultaneously satisfied.

The given equations are

5x+y=9    ......(1)

3x+2y=4  ......(2)

Here, the substitution method is used to solve the system of equations.

Find the value of y from equation (1) by subtracting 5x from both sides.

5x+y-5x=9-5x

y=9-5x

To find the value of x substitute the value of y in equation (2).

3x+2(9-5x)=4

Apply the distributive property a(b+c)=ab+ac as

3x+2×9-2×5x=4

3x+18-10x=4

Combine the like terms on the left side as

-7x+18=4

Subtract 18 from both sides and get

-7x+18-18=4-18

-7x=-14

Divide both sides by -7 and get

(-7x)÷(-7)=(-14)÷(-7)

x=2

Substitute the value of x in equation (1) and get

5(2)+y=9

10+y=9

Subtract 10 from both sides

10+y-10=9-10

y=-1

Hence, the solution of system of equations 5x+y=9 and 3x+2y=4 is (2,-1).

Learn more about system of equations from here brainly.com/question/13729904

#SPJ4

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Using differential calculus, maximize the volume of a box made of cardboard (top is open) as shown in Figure A. 15, subject to t
lana [24]

The maximum volume of the box is 40√(10/27) cu in.

Here we see that volume is to be maximized

The surface area of the box is 40 sq in

Since the top lid is open, the surface area will be

lb + 2lh + 2bh = 40

Now, the length is equal to the breadth.

Let them be x in

Hence,

x² + 2xh + 2xh = 40

or, 4xh = 40 - x²

or, h = 10/x - x/4

Let f(x) = volume of the box

= lbh

Hence,

f(x) = x²(10/x - x/4)

= 10x - x³/4

differentiating with respect to x and equating it to 0 gives us

f'(x) = 10 - 3x²/4 = 0

or, 3x²/4 = 10

or, x² = 40/3

Hence x will be equal to 2√(10/3)

Now to check whether this value of x will give us the max volume, we will find

f"(2√(10/3))

f"(x) = -3x/2

hence,

f"(2√(10/3)) = -3√(10/3)

Since the above value is negative, volume is maximum for x = 2√(10/3)

Hence volume

= 10 X 2√(10/3)  -  [2√(10/3)]³/4

= 2√(10/3) [10 - 10/3]

= 2√(10/3) X 20/3

= 40√(10/27) cu in

To learn more about Maximization visit

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Complete Question

(Image Attached)

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