Solve by substitution or elimination method:
4x – 5y = 14
-12x + 15y = -42
when i try this solution i end up cancelling out the whole problem. can someone please help me!
that's because you are really given only one equation
if you multiply the first equation by -3 you get your second equation so by subtracting them you get
0 = 0
if that happens, there will be an infinite number of solutions, namely any point that satisfies the first equation.
what about this one? I'm having a tough time with this one as well...
Solve by substitution or elimination method: -2x + 6y = 19
10x – 30y = -15
with the first one was i suppose to multiply it by -3? i multiplied the first equiation by 15 and the second by -5. does it matter?
SUBSTITUTION: Solve for a variable in one equation and then substitute that value for the variable in the other equation.
ELIMINATION: Manipulate an equation so that one variable will be eliminated when you add the equations together. As long as one variable cancels, it doesn't matter how you choose to manipulate the equation (as long as you do the same things to all terms in the equation).
so can you tell me if this is correct....
4x – 5y = 14
-12x + 15y = -42
15(4x – 5y = 14) = 60x – 75y = 210
5(-12x +15y = -42) = -60x + 75y = -210
60x – 75y = 210
+ -60x + 75y = -210
= 0
The equations are not independent therefore they have an infinite solution set.
would it be correct to say that the equations are not independent therefore they have an infinite solution set.
No. They are DEPENDENT (the lines are right on top of each other when you graph them) so they have an infinite solution set.
INDEPENDENT means that there's one point that's independent of the others (the point of intersection or the single solution to both equations).
Oh, you said NOT independent. I missed the double negative there - sorry.
To solve the system of equations using the substitution or elimination method, we'll start by showing you how to solve it using the elimination method.
Given the system of equations:
1) 4x - 5y = 14
2) -12x + 15y = -42
To eliminate one of the variables, we need to find a way to make the coefficients of either x or y in both equations equal. Let's eliminate the x variable.
To do that, we'll multiply equation 1 by 3 and equation 2 by 1 to adjust the coefficients of x:
1) (3)(4x - 5y) = (3)(14) --> 12x - 15y = 42
2) (1)(-12x + 15y) = (1)(-42) --> -12x + 15y = -42
Now, we have:
3) 12x - 15y = 42
4) -12x + 15y = -42
By adding equations 3 and 4 together, the x variable will be eliminated:
(12x - 15y) + (-12x + 15y) = 42 + (-42)
0 = 0
It seems like you've correctly noticed that the x variable cancels out, resulting in 0 = 0. This means that the two original equations are equivalent and consistent, indicating that they represent the same line in a graph. Therefore, there is no unique solution to this system of equations.
In terms of substitution, it won’t lead to a different outcome. You will still end up with an equation similar to 0 = 0, indicating an infinitely many solutions scenario.
Hence, the system of equations is dependent, and the solution consists of all the points on the line represented by the equations.