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3 years ago

What is the solution to the linear system x=-6y+15 and -2y+5x =11

Mathematics

1 answer:

Cerrena [4.2K]3 years ago

7 0

After solving the linear system, value of x=3 and value of y=2

Step-by-step explanation:

We need to find the solution to the linear system $x=-6y+15\,\,and\,\, -2y+5x =11$

$x=-6y+15\,\,\,\,(1)\\-2y+5x =11\,\,\,\,(1)$

Putting value of equation (1) x=-6y+15 in equation(2)

$-2y+5x =11\\-2y+5(-6y+15)=11\\-2y-30y+75=11\\-32y=11-75\\-32y=-64\\y=\frac{-64}{-32}\\y=2$

value of y=2

Now putting the value of y in equation (1)

$x=-6y+15\\x=-6(2)+15\\x=-12+15\\x=3$

value of x = 3

After solving the linear system, value of x=3 and value of y=2

Keywords: Linear system

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2 years ago

The graph shows the feasible region for the system with constraints:

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Answer:

The vertices feasible region are (0 , 15) , (10 , 15) , (20 , 5)

The minimum value of the objective function C is 125

Step-by-step explanation:

* Lets look to the graph to answer the question

- There are 3 inequalities

# y ≤ 15 represented by horizontal line (purple line) and cut the

y-axis at point (0 , 15)

# x + y ≤ 25 represented by a line (green line) and intersected the

x-axis at point (25 , 0) and the y- axis at point (0 , 25)

# x + 2y ≥ 30 represented by a line (blue line) and intersected the

x-axis at point (30 , 0) and the y-axis at point (0 , 15)

- The three lines intersect each other in three points

# The blue and purple lines intersected in point (0 , 15)

# The green and the purple lines intersected in point (10 , 15)

# The green and the blue lines intersected in point (20 , 5)

- The three lines bounded the feasible region

∴ The vertices feasible region are (0 , 15) , (10 , 15) , (20 , 5)

- To find the minimum value of the objective function C = 4x + 9y,

substitute the three vertices of the feasible region in C and chose

the least answer

∵ C = 4x + 9y

- Use point (0 , 15)

∴ C = 4(0) + 9(15) = 0 + 135 = 135

- Use point (10 , 15)

∴ C = 4(10) + 9(15) = 40 + 135 = 175

- Use point (20 , 5)

∴ C = 4(40) + 9(5) = 80 + 45 = 125

- From all answers the least value is 125

∴ The minimum value of the objective function C is 125

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