GRAPH FROM SLOPE INTERCEPT EQUATION

Graph the following equations of lines :

(Hint : Determine the slope and y – intercept)

Problem 1 :

y = -4x + 5

Solution :

y = -4x + 5

By comparing the given equation with slope intercept form, we get

y = mx + b

m = -4

So, slope is -4 and y– intercept is 5.

Problem 2 :

y = (3/4)x - 1

Solution :

y = (3/4)x - 1

y = mx + b

m = 3/4

So, the slope is 3/4 and y – intercept = -1.

Problem 3 :

y = 2x - 2

Solution :

y = 2x - 2

y = mx + b

m = 2

So, the slope is 2 and y – intercept is -2.

Problem 4 :

y = (-5/2)x + 3

Solution :

y = (-5/2)x + 3

m = -5/2

So, the slope is -5/2 and y – intercept is 3.

Problem 5 :

y = -4x

Solution :

y = -4x

y = mx + b

m = -4

So, the slope is -4 and y – intercept is 0.

Problem 6 :

y = 3x – 3

Solution :

y = 3x – 3

m = 3

So, the slope is 3 and y – intercept is -3.

Problem 7 :

Match the system of linear equations with the corresponding graph. Use the graph to estimate the solution. Check your solution

Solution :

Problem 1 :

y = 2.5x + 1 ------(1)

y = x ------(2)

Comparing the given equations with y = mx + b

Slope (m₁) = 2.5 and y-intercept (b₁) = 1

Slope (m₂) = 1 and y-intercept (b₂) = 0

Both should be raising line.

Problem 2 :

y = 2x - 3 ------(1)

y = (-1/2)x + 1 ------(2)

Comparing the given equations with y = mx + b

Slope (m₁) = 2 and y-intercept (b₁) = -3

Slope (m₂) = -1/2 and y-intercept (b₂) = 1

The first one should be raising line and the second line should be falling line.

Problem 3 :

y = (1/4)x - 2 ------(1)

y = (-2/5)x + 4 ------(2)

Slope (m₁) = 1/4 and y-intercept (b₁) = -2

Slope (m₂) = -2/5 and y-intercept (b₂) = 4

The first one should be raising line and the second line should be falling line.

Option A : 

y-intercept of falling line = 4

y-intercept of raising line = -3

Matches with problem 2.

Option B : 

y-intercept of falling line = 1

y-intercept of raising line = 0

Matches with Problem 1

Option C : 

y-intercept of falling line = 4

y-intercept of raising line = -2

Matches with problem 3

Problem 8 :

Solve the equations algebraically and graphically.

(1/2)x - 2 = (-1/4)x + 1

Solution :

Solving the equation algebraically :

(1/2)x - 2 = (-1/4)x + 1

(1/2) x + (1/4)x = 1 + 2

(2x + x)/4 = 3

3x/4 = 3

x = 3(4/3)

x = 4

Applying the value in y = (1/2)x - 2

y = (1/2)(4) - 2

y = 2 - 2

y = 0

So, the point of intersection is (4, 0). By observing the graph, we get the points of intersection is (4, 0).

Problem 9 :

Solve the equations algebraically and graphically.

(2/3)x + 4 = (1/6)x + 5/2

Solution :

Solving the equation algebraically :

(2/3)x + 4 = (1/6)x + 5/2

(2/3) x - (1/6)x = (5/2) - 4

(4x - x)/6 = (5 - 8)/2

3x/6 = -3/2

x/2 = -3/2

x = -3

Applying the value in y = (2/3)x + 4

y = (2/3)(3) + 4

y = 2 + 4

y = 6

So, the point of intersection is (-3, 6). By observing the graph, we get the points of intersection is (-3, 6).

Problem 10 :

You have 4x + 10 CDs and your friend has 5x + 22 CDs. Is it possible that you both have the same number of CDs? Explain your reasoning.

Solution :

4x + 10 = 5x + 22

4x - 5x = 22 -10

-x = 12

x = -12

Number of CDs cannot be negative. So, the situation is not possible.