Answer :
so y=mx+b
the 'm' part means the slope which is found by 'rise/run'
a. y=2x+3 or y=2x+3
b. y=-1x+6
c. i'm not sure what this part means
d. this means 'what values will work for x and y in both equations at once so
y=-x+6 and y=2x+3
y=-x+6
add x to both sides
x+y=6
subtitute 'y=2x+3' for y in x+y=6
x+(2x+3)=6
collect like terms
3x+3=6
subtract 3 from both sides
3x=3
divide both sides by 3
x=1
sibtitue into x+y=6
1+y=6
subtract 1 from both sides
y=5
(1,5) is the solution (x,y)
Question 2
so amount of peanuts (x) =60%
ammount of almonds (y)=40%
y=2(1/3x) or y=2/3x
so there is a mixture that is8/10 almonds
for
a. i'm not sure about this one
b. so 60% of 5 pounds=3 pounds
40% of 5 pounds=2pounds
we need to find out how many pounds, times 8/10, = 2pounds
she will need 2.5 pounds of the pure peanuts and 2.5 pounds of the almond-peanut mixture
the 'm' part means the slope which is found by 'rise/run'
a. y=2x+3 or y=2x+3
b. y=-1x+6
c. i'm not sure what this part means
d. this means 'what values will work for x and y in both equations at once so
y=-x+6 and y=2x+3
y=-x+6
add x to both sides
x+y=6
subtitute 'y=2x+3' for y in x+y=6
x+(2x+3)=6
collect like terms
3x+3=6
subtract 3 from both sides
3x=3
divide both sides by 3
x=1
sibtitue into x+y=6
1+y=6
subtract 1 from both sides
y=5
(1,5) is the solution (x,y)
Question 2
so amount of peanuts (x) =60%
ammount of almonds (y)=40%
y=2(1/3x) or y=2/3x
so there is a mixture that is8/10 almonds
for
a. i'm not sure about this one
b. so 60% of 5 pounds=3 pounds
40% of 5 pounds=2pounds
we need to find out how many pounds, times 8/10, = 2pounds
she will need 2.5 pounds of the pure peanuts and 2.5 pounds of the almond-peanut mixture
