Find an equation of the line whose intercepts are twice those of the graph of 5y+2x=10.

X intercept is where the line crosses the x axis or where y=0

y intercept is where the line crosses the y axis or where x=0

so find x and y intercept

yintercept, x=0

5y=10

y=2

yint is (0,2)

xint, y=0

2x=10

x=5

xint is (5,0)

double

yint is (0,4) and yint is (10,0)

ok, so we do

y=mx+b

m=slope

b=yint

givenyint=4

y=mx+4

when x=10, y=00 because we got (10,0)

0=10m+4

miinus 4 both sides

-4=10m

divide oth sides by 10

-2/5=m

the equation is y=-2/5x+4 or

2/5x+y=4

2x+5y=20

5y+2x=10 y = - 25 x+2 2 4 6 8 10 - 2 - 4 - 6 - 8 - 10 2 4 6 8 10 - 2 - 4 - 6 - 8 - 10 Let's solve for x. 5y + 2x = 10 Step 1: Add - 5y to both sides. 2x + 5y + - 5y = 10 + - 5y 2x = - 5y + 10 Step 2: Divide both sides by 2. 2x 2 = - 5y + 10 2 x = - 5 2 y + 5 Answer: x = - 5 2 y + 5