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Find values for m and b in the following system so that the solution to the systemis (-3, 4).
5x + 7y = b
mx + y = 22

2006-08-22 05:04:55 · 17 answers · asked by sistermoon 4 in Education & Reference Homework Help

17 answers

(-3, 4) is a point (x,y)

sub -3 in for x & 4 in for y

5(-3)+7(4)=b
-15+28=b
13=b

m(-3)+(4)=22
-3m=18
m=-6

so b (your constant) is 13 & m (your y-intercept) is -6

2006-08-22 05:13:45 · answer #1 · answered by h-e-a-t-h-e-r 3 · 0 0

the answer to this is simple
put the values of x, y in the equation so as to have just two variables m,b

putting the values in eqn 1 :

5(-3) + 7(4) = b
m(-3) + 4 = 22

thus b = 13 and m = -6

got that

2006-08-22 12:16:56 · answer #2 · answered by iruleusmntp 1 · 0 0

-3 = x and 4 = y

Just substitute them into the equations

5(-3) + 7(4) = b
-15+28 = b
13 = b

and

m(-3) + 4 = 22
-3m = 22-4
-3m = 18
m = 18/(-3)
m = -6

2006-08-22 12:12:17 · answer #3 · answered by jimvalentinojr 6 · 0 0

to find the B substite (-3,4) in the first equation

5(-3)+ 7(4)=b
-15 + 28=b
13= b

to find m sub. (-3,4) in the secomd equation
(-3)m +(4)= 22
-4 -4
-3m=18
dived both sides by -3
m= -6

b=13 and m= - 6



later you would have to k now:
a linar exation y=mx+b
where m is the slope and b is the y intercept, and the (x,y) are any points on the line

2006-08-22 12:11:06 · answer #4 · answered by 3umar 3 · 0 0

-5*3+7*4=b
-3m +4=22
{b=13
{m= -6

2006-08-22 12:16:28 · answer #5 · answered by Anonymous · 0 0

B=13
M=-6

2006-08-22 12:10:34 · answer #6 · answered by gurusenthilkumar g 2 · 0 0

5*(-3) + 7*4=b
-15+28=b
b=13

m*(-3) + 4=22
-3m=18
m=-6

2006-08-22 12:12:45 · answer #7 · answered by giviko2 2 · 0 0

5(-3)+7(4)=b
2+11=b
13=b

m(-3)+4=22
-4 -4
-3m = 18
-3 -3
m=-6

2006-08-22 12:15:44 · answer #8 · answered by Anonymous · 0 0

m(-3) + (4) = 22
-3m = 18
m = 18/-3
m = -6

5(-3) + 7( 4) = b
-15 + 28 = b
13 = b

2006-08-22 12:11:39 · answer #9 · answered by smartee 4 · 0 0

Just plug in -3 for the x and 4 for the y then solve!

2006-08-22 12:11:33 · answer #10 · answered by Anonymous · 0 0

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