This is number a and I will list take the other
we have point (-6, - 1)
Now we will put these points in each equation,
y = 4x +23
put x = -6 and y = -1
-1 = 4 (-6) +23
-1 = -24 + 23
-1 = -1
LHS = RHS, so this equation has (-6 , -1) as solution.
y = 6x
put x = -6 and y = -1
-1 = 6 (-6)
-1 not= -36
LHS is not equal RHS, so (-6 , -1) is not a solution for that equation,
y = 3x - 5
put x = -6 and y = -1
-1 = 3 (-6) - 5
-1 = -18 - 5
-1 not= -23
LHS is not equal RHS, so (-6 , -1) is not a solution for that equation,
y= 1/6 x
put x = -6 and y = -1
-1 = -6/6
-1 = -1
LHS = RHS, so (-6 , -1) is a solution for that equation,
It looks like your equations are
7M - 2t = -30
5t - 12M = 115
<u>Solving by substitution</u>
Solve either equation for one variable. For example,
7M - 2t = -30 ⇒ t = (7M + 30)/2
Substitute this into the other equation and solve for M.
5 × (7M + 30)/2 - 12M = 115
5 (7M + 30) - 24M = 230
35M + 150 - 24M = 230
11M = 80
M = 80/11
Now solve for t.
t = (7 × (80/11) + 30)/2
t = (560/11 + 30)/2
t = (890/11)/2
t = 445/11
<u>Solving by elimination</u>
Multiply both equations by an appropriate factor to make the coefficients of one of the variables sum to zero. For example,
7M - 2t = -30 ⇒ -10t + 35M = -150 … (multiply by 5)
5t - 12M = 115 ⇒ 10t - 24M = 230 … (multiply by 2)
Now combining the equations eliminates the t terms, and
(-10t + 35M) + (10t - 24M) = -150 + 230
11M = 80
M = 80/11
It follows that
7 × (80/11) - 2t = -30
560/11 - 2t = -30
2t = 890/11
t = 445/11
Y = 3x - 4
y = 3(-1) - 4
y = -3 - 4 = -7
Graph pair is (-1, -7)
y = 3(0) - 4
y = 0 - 4 = -4
Graph pair is (0, -4)
y = 3(2) - 4
y = 6 - 4 = 2
Graph pair is (2, 2)
y = 3(4) - 4
y = 12 - 4 = 8
Graph pair is (4, 8)
Answer:
r(t) = 15t+500
Step-by-step explanation:
Since the amount of money (r) is a function of the time (t) we will make it the y-value. t is how much time and he gets $15 a minute so we multiply t by 15. 500 is how much he gets paid for doing it. If he showed up and just left, he would still get 500.
