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I know the answer, I just need help wording it. Help me out? (Will fan + medal) -------- Consider the following pair of equations: y = −2x + 8 y = x − 1 Explain how you will solve the pair of equations by substitution. Show all the steps and write the solution in (x, y) form. ----- I know the answer is (3, 2) I just need help with writing the equation. I'm not good with words

Mathematics
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I know that since we already have what y equals on both equations, you combine them and solve. I just need help writing down what I did in my mind.
start of by making x-1 = -2x+8
in words that would be " first equation equals the second"

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Other answers:

@alekos Yup! I got that written down to make the equations equal to each other.
We can plug in x - 1 into the first equation then solve for 'x'..after we get 'x' we plug it back into any of the two equations to find 'y'. Note that the solutions are in (x, y) form.
The question says: Explain how you will solve the pair of equations by substitution. Show all the steps and write the solution in (x, y) form. @alekos is correct in saying x-1 = -2x +8 but that doesnt fully answer the question.
\(\sf y = -2x + 8\) \(\sf y = x - 1\) \(\sf x - 1 = -2x + 8\_ Add 2x to both sides: \(\sf 3x - 1 = 8\) Add 1 to both sides: \(\sf 3x = 9\) Divide 3 to both sides: \(\sf x = 3\) Now plug this into any of the two equations to find 'y': \(\sf y = x - 1\) \(\sf y = 3 - 1\) \(\sf y = 2\) So our solution is \(\sf (3, 2)\).
"in words" @iGreen
Not sure what happened there.. \(\sf x - 1 = -2x + 8\) Add 2x to both sides: \(\sf 3x - 1 = 8\)
@BPDlkeme234 "We can plug in x - 1 into the first equation then solve for 'x'..after we get 'x' we plug it back into any of the two equations to find 'y'."
Yes, @iGreen is explaining in words!
I said that earlier..lol.
Ooops! sorry @iGreen
You start with\[x-1=-2x+8\]Then you subtract 8 from both sides x - 1= -2x + 8 -8 -8 -9 Then you divide 2 from both sides -9/-2 \[x=4.5\]
Can I get what you said?
@Study_together this has already been solved by @iGreen
More than 1 person can do it \(\Huge\ddot\smile\)

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