1. 3x-2y=24
2. 6x-4y=-6
Note that we can write the first equation as -2y=-3x+24
and the second as -4y=-6x-6
Now divide the first equation by -2 and the second by -4
1. y=3/2x-12
2. y=3/2x+6/4
This tells us the lines have the same slope and different y intercepts which means they are parallel. There is NO solution.
If you use elimination or substitution you find the same thing.
Note, if you try emiminate you end up with 0y=-54 which is not true also telling you there is no solution.
How many solutions are there to the following system of equations?2x - y = 2-x + 5y = 3if this is your question,there is ONLY 1 way to solve it.
You solve equations with fractions the same way you solve other equations. You perform various arithmetic operations on both sides of the equals sign until you get the result you want.
To solve for two unknown variables (x and y) you require two independent equations,
why will the equations x+14=37 and x-14=37 have different solutions for x
It is not possible to solve one equation in two unknowns (x and y). Two independent equations are required.
You can use a graph to solve systems of equations by plotting the two equations to see where they intersect
x = -1.2, y = -3
How many solutions are there to the following system of equations?2x - y = 2-x + 5y = 3if this is your question,there is ONLY 1 way to solve it.
If you already know that x = -3 and y = 5 what linear equations are you wanting to solve?
You solve equations with fractions the same way you solve other equations. You perform various arithmetic operations on both sides of the equals sign until you get the result you want.
The answer depends on whether they are linear, non-linear, differential or other types of equations.
because you need maths in your life.. everyone does
solve systems of up to 29 simultaneous equations.
The solutions are: x = -2 and y = 4
To solve a system of equations, you need equations (number phrases with equal signs).
Equals divided by non-zero equals are equal.
You would solve them in exactly the same way as you would solve linear equations with real coefficients. Whether you use substitution or elimination for pairs of equations, or matrix algebra for systems of equations depends on your requirements. But the methods remain the same.