Will Medal and Fan What is the solution to the system of equations? 5x-3y=-9 2x-5y=4
@One098 @Opcode @wio @TheAsker2002
@xo.minnie.xox
Can you isolate a variable in the first equation?
What I mean is, for: \[ 5x-3y=9 \]Can you solve for \(x\) or \(y\)? Your first step is to solve for one of them.
5x-3y=9 x=3 and y=2
@wio
Well, that would be one solution, but we don't want to find a single solution for it, because it might not be the solution for the other equation.
ohh ok
Suppose we want to isolate \(x\). First we would add \(3y\) to both sides: \[5x = 9+3y\]Then, we would divide by \(5\): \[ x= \frac{9}{5}+\frac {3y}{5}\]
We can substitute this for the \(x\) in the other equations:\[ 2x-5y=4 \]Will become: \[ 2\left(\frac 95+\frac {3y}5\right)-5y=4 \]Now we can solve for \(y\).
\[ \frac{18}5+\frac{6y}5-5y = 4 \]We should multiply everything by \(5\) to get rid of fractions: \[ 18+6y-25y=20 \]
18+6y-25y=20 so every number in this equation multiplied by 5
Hold on one second...
I made a mistake, I was using\[ 5x-3y=\color{red}9 \]But I should be using \[ 5x-3y=\color{red}{-9} \]So, fixing that mistake results in: \[ -18+6y-25y=4 \]
Now, we can combine the like terms for \(y\) to get: \[ -18-19y = 20 \]
ohhh ok is that our last step to the equation?
Then we add \(18\): \[ -19y = 38 \]And this gives us \(y=-2\). Does that make sense?
Now, to solve \(x\), we substitute our solution for \(y\).\[ 5x-3(-2)=-9 \]
Can you solve for \(x\)?
yes thank you for not giving me the answer and explaning :)
@wio how does that go to my answer choices A. (0, 4) B. (–4, 0) C. (0, –4) D. (4, 0)
Sorry, but those answer choices can't be for the same question.
Use the graph to state the solution for the system. x – y = 4 (line a) x + y = 4 (line b) A. (0, 4) B. (–4, 0) C. (0, –4) D. (4, 0)
The answer is \((x,y) = (-3,-2)\).
Well, this is a separate question, but the point where they intersect is the solution.
And they appear to intersect at \((4,0)\).
Thank you for ur time! : )
im gonna open another question can you help me with that one to @wio
Sorry, but I have to go soon.
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