Question

Write a linear equation for the tables shown. y = mx + b y =__ x +____

x y x y
-6 12 -14 3
-5 14 -12 6
-4 16 -10 9
-3 18 - 8 12
-2 20 - 6 15

Answer 1

ANSWER TO QUESTION 1

$y=2x+24$

EXPLANATION

Method 1: Finding the equation given any 2 points

We  choose any two of the ordered pairs from the first table, say

$(-2,20)$

and

$(-3,18)$

We determine the slope using the formula;

$Slope=\frac{y_2-y_1}{x_2-x_1}$

Let $(x_1,y_1)$ be $(-2,20)$

and

$(x_2,y_2)$ be $(-3,18)$

Then,

$Slope=\frac{18-20}{-3--2}$

$\Rightarrow Slope=\frac{18-20}{-3+2}$

$\Rightarrow Slope=\frac{-2}{-1}$

$\Rightarrow Slope=2$

We now use  the formula,

$y-y_1=m(x-x_1)$ to find the equation of this line.

That is ;

$y-20=2(x--2)$

$y-20=2(x+2)$

$y=2x+4+20$

$y=2x+24$

ANSWER TO QUESTION 2

Method 2: Using Simultaneous equations

We use the slope intercept form  for the second table

$y=mx+b$

The point

$(-6,15)$ must satisfy this line.

$15=-6m+b--(1)$

The point

$(-8,12)$ must also satisfy this line.

$12=-8m+b--(2)$

Equation (1)  minus Equation (2) gives

$2m=3$

$\Rightarrow m=\frac{3}{2}$

We substitute $m=\frac{3}{2}$ in to equation (1)

and solve for b.

$\Rightarrow 15=-6\times \frac{3}{2}+b$

$\Rightarrow 15=-9+b$

$\Rightarrow 15+9=b$

$\Rightarrow 24=b$

Solving simultaneously gives

$b=24$ and $m=\frac{3}{2}$

Hence the equation is

$y=\frac{3}{2}x+24$

Answer 2

Answers:

1) First table: y=2x+24

2) Second table: y=(3/2)x+24

Solution:

m=(y2-y1)/(x2-x1)

y-y1=m(x-x1)

1) First table. Taking the first two points of the table:

P1=(-6,12)=(x1,y1)→x1=-6, y1=12

P2=(-5,14)=(x2,y2)→x2=-5, y2=14

m=(y2-y1)/(x2-x1)

m=(14-12)/(-5-(-6))

m=(2)/(-5+6)

m=(2)/(1)

m=2

y-y1=m(x-x1)

y-12=2(x-(-6))

y-12=2(x+6)

y-12=2x+2(6)

y-12=2x+12

y-12+12=2x+12+12

y=2x+24

2) Second table. Taking the first two points of the table:

P1=(-14,3)=(x1,y1)→x1=-14, y1=3

P2=(-12,6)=(x2,y2)→x2=-12, y2=6

m=(y2-y1)/(x2-x1)

m=(6-3)/(-12-(-14))

m=(3)/(-12+14)

m=(3)/(2)

m=3/2

y-y1=m(x-x1)

y-3=(3/2)(x-(-14))

y-3=(3/2)(x+14)

y-3=(3/2)x+(3/2)(14)

y-3=(3/2)x+(3)(14)/2

y-3=(3/2)x+42/2

y-3=(3/2)x+21

y-3+3=(3/2)x+21+3

y=(3/2)x+24