Answer:
x=133 y=-25
Step-by-step explanation:
I'll do both ways for you. So let's start with Substitution:
With the sub method, you have to set both equations equal to each other by setting them equal to the same variable. Since there is no coefficient in front of both x's in both equations, that variable will be easiest to solve for.
x + 2y = 83 & x + 5y = 8
Solve for x.
x = 83 - 2y & x = 8 - 5y
Once you have solved for x, set each equation equal to one another and solve for y now.
83 - 2y = 8 - 5y
Isolate all variables to one side:
83 = 8 - 3y
Now subtract the 8 to fully isolate the y variable:
75 = -3y
Divide by -3:
-25 = y Now that you have your first variable, plug it into one of the original equations and solve for x.
x + 2(-25) = 83
x - 50 = 83
x = 133
Now for the Elimination method. For this method you need to get rid of a variable by either subtracting/adding each equation together. Again, since you can subtract and x from both equations, you will be left with only the y variable to solve:
Put each equation on top of one another and subtract:
x + 2y = 83
- (x + 5y = 8)
The x's will cancel out:
(x - x) + (2y - 5y) = (83 - 8)
Simplify:
-3y = 75
Solve for y
y = -25
Then, plug y = -25 into one of the original equations:
x + 5(-25) = 8
Solve for x:
x - 125 = 8
x = 133
Hope this helps!
We can start from the given line's coefficients and translate the line from the origin to the given point.
4(x -(-2)) -(y -3) = 0
4x +8 -y +3 = 0
The equation of the desired line is ...
4x -y = -11
_____
For standard form line ax+by=c, any parallel line will have only a different value of c. For c=0, the line goes through the origin (0, 0). To make it go through point (h, k) we can write it as
a(x-h) +b(y-k) = 0
which is completely equivalent to
ax +by = ah +bk
Answer:
what do you want to be answered
Step-by-step explanation:
Let's represent the two numbers by x and y. Then xy=60. The smaller number here is x=y-7.
Then (y-7)y=60, or y^2 - 7y - 60 = 0. Use the quadratic formula to (1) determine whether y has real values and (2) to determine those values if they are real:
discriminant = b^2 - 4ac; here the discriminant is (-7)^2 - 4(1)(-60) = 191. Because the discriminant is positive, this equation has two real, unequal roots, which are
-(-7) + sqrt(191)
y = -------------------------
-2(1)
and
-(-7) - sqrt(191)
y = ------------------------- = 3.41 (approximately)
-2(1)
Unfortunately, this doesn't make sense, since the LCM of two numbers is generally an integer.
Try thinking this way: If the LCM is 60, then xy = 60. What would happen if x=5 and y=12? Is xy = 60? Yes. Is 5 seven less than 12? Yes.
