SOLUTION: I am homeschooled with the School of Tomorrow. I am currently doing Algebra 1, and am having a lot of touble with the section "Solving Simultaneous Equations Algebraically", lesson

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Question 197661: I am homeschooled with the School of Tomorrow. I am currently doing Algebra 1, and am having a lot of touble with the section "Solving Simultaneous Equations Algebraically", lesson "Substitution Method." The equation I am trying to work out now is:
3x-4y=10
x=6y-1
Found 2 solutions by jim_thompson5910, checkley77:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
3x-4y=10 Start with the first equation.


3%286y-1%29-4y=10 Plug in x=6y-1 into the first equation. In other words, replace each x with 6y-1. Notice we've eliminated the x variables. So we now have a simple equation with one unknown.


18y-3-4y=10 Distribute


14y-3=10 Combine like terms on the left side


14y=10%2B3Add 3 to both sides


14y=13 Combine like terms on the right side


y=13%2F14 Divide both sides by 14.


Now that we know that y=13%2F14, we can plug this into x=6y-1 to find x



x=6%2813%2F14%29-1 Substitute 13%2F14 for each y


x=78%2F14-1 Multiply


x=39%2F7-1 Reduce


x=32%2F7 Combine like terms.


So our answers are x=32%2F7 and y=13%2F14 which form the ordered pair


Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
3x-4y=10
x=6y-1
now replace x in the first equation with(6y-1) & solve for y.
3(6y-1)-4y=10
18y-3-4y=10
14y=10+3
14y=13
y=13/14 ans.
x=6*13/14-1
x=78/14-1
x=(78-1*14)/14
x=(78-14)/14
x=64/14
x=32/7 ans.
Proof:
3*32/7-4*13/14=10
96/7-52/14=10
(96*2-52)/14=10
(192-52)/14=10
140/14=10
10=10