Consider the following system of equations. Solve and write as a coordinate point.
5x+4y=-14
3x+6y=6
Please help and thanks
You have to decide which variable to eliminate.
Lets choose x. We need the coefficients to be equal and opposite. I will multiply the first equation by 3 and the second equation by -5
15x + 12y =42
-15x -30y = -30
Add the two equations and the x will be eliminated.
-18y = 12 y = -12/18 = -2/3
To find x - replace y with the answer. I will choose the second equation because it seems easier.
3x + 6(-2/3) = 6
3x -4 = 6
3x = 10
x = 10/3
(10/3, -2/3)
Check in both equation to be sure that a mistake hasn't been made.
well, I think I would multiply the first one by three and the second one by two
15 x + 12 y = -42
6 x + 12 y = 12
------------------ subtract
9 x = -54 or x = -6
then
-18 + 6 y = 6
6 y = 24
y = 4
(-6 , 4 )
John's error, after multiplying by 3 , should be
15x + 12y = -42 , John had +42
To solve this system of equations, you can use the method of substitution or elimination. I will explain how to solve it using the method of elimination:
Step 1: Multiply one of the equations by a coefficient to make the coefficients of either x or y the same in both equations. In this case, let's multiply the second equation by 2 to make the coefficients of y the same.
The equations become:
5x + 4y = -14
6x + 12y = 12
Step 2: Now, subtract one equation from the other to eliminate the y variable. Let's subtract the first equation from the second equation.
(6x + 12y) - (5x + 4y) = 12 - (-14)
x + 8y = 26
Step 3: Now we have one equation with only the variable x. We can solve it for x.
x = 26 - 8y
Step 4: Substitute the value of x back into one of the original equations to solve for y. Let's use the first equation.
5(26 - 8y) + 4y = -14
130 - 40y + 4y = -14
-36y = -144
y = -144 / -36
y = 4
Step 5: Substitute the value of y back into the equation we found in step 3 to solve for x.
x = 26 - 8(4)
x = 26 - 32
x = -6
Therefore, the solution to the system of equations is (x, y) = (-6, 4).