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algebra 2 word problems?

for 30 minutes you do a combinations of walking and jogging. at the end of your workout your pedometer displays a total of 2.5 miles. you know that you walk 0.05 mile per minute and jog 0.1 mile per minute. for how much time were you walking? for how much time were you jogging? use a verbal model to write and solve a system of linear equations.

where x equals time spent walking and y equals time spent jogging
0.05x+.1y=2.5 and x+y=30
there are two possible ways to solve this

1 substitution:
x+y=30
x=30-y
then replace x in .05x+.1y=2.5 to get .05(30-y)+.1y=2.5
solve as follows:
.05(30-y)+.1y=2.5
1.5-.05y+.1y=2.5
1.5+.05y=2.5
.05y=2.5-1.5
.05y=1
y=(1/.05)
y=20
replace y=20 in your original equation .05x+.1y=2.5 and solve for x
.05x+.1(20)=2.5
.05x+2=2.5
.05x=2.5-2
.05x=.5
x=(.5/.05)
x=10

2 elimination
take the same 2 equations and zero out either x or y. in my example i will zero out y
multiply .05x+.1y=2.5 by -10
.05x+.1y=2.5 becomes -.5x-1y=-25
x+y=30 becomes x+y=30
subtract one equation from the other
x+y=30
-.5x-y=-25
.5x=5
x=(5/.5)
x=10 insert into either one of you original equations. i will use x+y=30 simply because its easiest
10+y=30
y=(30-10)
y=20
x=10 and y=20 so the time spent walking was 10 minutes and the time spent jogging was 20 minutes

i cant graph the equations for the second question but ill do my best to explain it
its simple really
store 1 gives a 15% discount which equates to selling the camera for an 85% cost of the original price. and the second store gives a flat $300 discount. so let x= the original price of the camera and let y= the discounted price
then store 1 is x(.85)=y or 700(.85)=595
and store 2 is x-300=y or 700-300=400
so at 700 for the price of the camera store 2 is cheaper
to find out when store 1 is cheaper then
x(.85) then: (.85)x .85x-x<-300
-.15x<-300
x<2000
so when the price of the camera is above $2000 store 1 is cheapest
to graph this using the slope intercept form
store 1 is y=.85x
store 2 is y=x-300

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