Solve the following system of equations: x - 2y = 14 x + 3y = 9
you need to find x and y coordinates ?
(1, 12) (-1, -12) (12, -1) (12, 1)
\[x -2y = 14\]\[x+3y=9\] Do you know substitution or combination?
lets do y first and get rid of x
so change all signs for the bottom so therefore -x-3y=-9
ok im not really good at math
Substitution is easy here as you can quickly get \(x\) in terms of \(y\) with little work. Then you plug the expression for \(x\) in place of \(y\) in the other equation and solve it for the value of \(y\). Once you know \(y\), you plug it in to your substitution equation to get \(x\).
so both x's cancels out
then you are left with -5y=-5
giving you 1 so y equals one
then replace your answer with y so x-2(1)=14 then x-2=14
so what would be the answer
its 12 isnt it
wait y would equal -1 sorry
\[x-2y =14\]\[x-2y+2y = 14 + 2y\]\[x = 14 + 2y\] Now we take our other equation: \[x + 3y = 9 \]and replace \(x\) with \(14+2y\) \[(14+2y) + 3y = 9\] \[14 + 2y + 3y = 9\]\[14 + 5y = 9\]\[14-14+5y = 9-14\]\[5y = -5\]\[y = -1\] Now we put our value of \(y = -1\) back in the substitution equation to find \(x\): \[x = 14 + 2y \]\[x = 14 + 2(-1)\]\[x = 14 - 2\]\[x = 12\] Now, the most important step! Check the work to make sure it solves BOTH equations: \[x-2y = 14\]\[12-2(-1) = 14\]\[12 + 2 = 14\checkmark\] \[x+3y=9\]\[12+3(-1) = 9\]\[12 -3 = 9\checkmark\]
so then x-2(-1)=14 then x+2=14 then x=14-2 making x=12
so it would be C
It is entirely possible to get a set of "answers" which only satisfies some of the equations. For example, x = 0 and y = 3 would work in \[x+3y=9\] but will not work for \[x-2y=14\]
Solve the system of equations by substitution. What is the solution for x? 2x + y = 1 4x + 2y = -1
i need help
Okay, let's do the same thing. \[2x+y = 1\]\[4x+2y = -1\] Can you solve one of those equations to give you 1 variable in terms of the other? That means get 1 variable all alone on one side of the = sign, everything else on the other.
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