help me solve the following using the elimination method: 6x+7y+2=0 and 7x-6y-26=0
multiply the first by 6 and the second by 7, then add the new equations.
It should be rather straightforward from there.
help me understand how to solve this problem x+4y=17
First, if you have a question, it is much better to put it in as a separate post in <Post a New Question> rather than attaching it to a previous question, where it is more likely to be overlooked.
Second, what are you trying to solve? You cannot get the numerical values of x and y with only one equation.
x = 17 - 4y
y = (17 - x)/4
I hope this helps. If not, repost your question in clearer terms.
To solve the system of equations using the elimination method, we need to eliminate one variable by adding or subtracting the equations. Here's how to solve it step by step:
Step 1: Multiply the equations by appropriate constants to make the coefficients of one of the variables equal in magnitude but opposite in sign.
Multiply equation 1 by 7: 42x + 49y + 14 = 0
Multiply equation 2 by 6: 42x - 36y - 156 = 0
Step 2: Add the two equations together to eliminate the x term.
(42x + 49y + 14) + (42x - 36y - 156) = 0
Combine like terms: 84x + 13y - 142 = 0
Step 3: Now we have an equation in terms of only the y variable. So we can solve for y.
84x + 13y - 142 = 0
Rearrange the equation: 13y = -84x + 142
Divide the entire equation by 13: y = (-84/13)x + (142/13)
Step 4: Substitute the value of y obtained in step 3 back into any of the original equations to solve for x. Let's use equation 1.
6x + 7y + 2 = 0
Substitute y = (-84/13)x + (142/13): 6x + 7((-84/13)x + (142/13)) + 2 = 0
Simplify the equation: (54/13)x - (994/13) = 0
Add (994/13) to both sides: (54/13)x = (994/13)
Multiply both sides by 13/54 to isolate x: x = (994/13) * (13/54)
Simplify: x = 38/27
Step 5: Substitute the value of x obtained in step 4 back into y = (-84/13)x + (142/13):
y = (-84/13) * (38/27) + (142/13)
Simplify the expression: y = -8/3
Thus, the solution to the system of equations is x = 38/27 and y = -8/3.