Solve a system of linear equations?
Given the table of values what is the solution to this system of equations?
Equation 1: x (-3, -1, 1, 3)
y(9, 4, -1, -6)
Equation 2: x(-2, -1, 0, 2)
y(6.5, 4, 1.5, -3.5)
I'm really stuck this problem, if anybody could tell me how to do it. Not necessarily tell me the answer, but just how to solve it
"So you need to find a pair of points that could work for both equations."
Yes, but how do i find those pairs of points
- AnnLv 69 years agoFavorite Answer
(-3, -1, 1, 3)
(9, 4, -1, -6)
You can see that x increases, y decreases. So y = (1/k) x + c
1/k because that makes it inversely proportional.
y = (1/k) x + c
9 = (1/ k) -3 + c
4 = (1/k) -1 + c
Two equations, two unknowns, now is just a matter of solving the system.
9 = -3/ k + c ==> c = 9 +3/k Plug that in the second equation.
4 = (1/k) -1 + c
4 = - 1/k + c
4 = -1/k + 9 + 3/k
-5 = -1/k +3/k
-5 = 2/k
-5k = 2
k = -2/5
c = 9 + 3/ (-2/5)
c = 9 -15/2
c = 18/2 - 15/2
c = 3/2
Edit: I plugged in the coordinates (-3,9), (-1,4) (1,-1) and (3, -6) and they all worked on the y = -2.5 x +1.5 line that you get for k = -2/5 and c = 3/2.
9 = -2.5 (-3) + 1.5
9 = +7.5 +1.5
4 = -2.5 (-1) +1.5
4 = 2.5 +1.5
-1 = -2.5 (1) +1.5
-1 = -2.5 +1.5
-6 = -2.5 * 3 +1.5
-6 = -7.5 +1.5
So I think what you need to do is make the same process with equation 2's coordinates, find the line for equation 2 and then solve for the x and y that satisfy both.
- lammonsLv 44 years ago
A equipment of linear equations is two or extra linear equations that are being solved concurrently. frequently, a answer of a equipment in 2 variables is an ordered pair that makes the two equations genuine. In different words, it relatively is the place the two graphs intersect, what they have in difficulty-loose. So if an ordered pair is a answer to a minimum of one equation, yet not the different, then it is not a answer to the equipment. A consistent equipment is a equipment that has a minimum of one answer. An inconsistent equipment is a equipment that has no answer There are 3 techniques to unravel systems of linear equations in 2 variables: graphing substitution approach removing approach
- Anonymous9 years ago
The first answer is incorrect. They're not asking for an equation for the first line. They're asking for the solution to the system.
You could try finding the equation for both lines, then solve that system algebraically, but there's a much easier way to do this.
The first line contains the points (-3,9), (-1,4), (1,-1), and (3,-6). The second line contains the points (-2, 6.5), (-1,4), (0,1.5) and (2,-3.5). Since (-1,4) is common to both lines, then it must be the solution to the system. When you graph two linear equations in two dimensions, you get two lines that either intersect at one point (one unique solution), are parallel (no intersection and thus no solution) or they're both two ways of writing the same exact line (infinite number of solutions).
- Hal RoachLv 79 years ago
Notice that (-1,4) shows up in both tables. So that must be the solution, because it's a point that's a solution to both equations. There's no need to go out of your way to find out what the equations are.