How to find the solution (x,y)

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To find the solution (x,y) for the equations x - y = 1 and 2x + y = -4, one must identify the values of x and y that satisfy both equations simultaneously. The geometric interpretation involves finding the intersection point of the two lines represented by these equations. A suggested approach is to solve one equation for y in terms of x and substitute it into the other equation. However, there was a mistake in the calculations where the addition of the equations was incorrectly simplified. The correct method involves adding the equations to eliminate y, leading to the correct solution.
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Homework Statement



The solution (x,y) of the system equations define by x-y=1 and 2x+y= -4 is what

Homework Equations





The Attempt at a Solution



I found x and y intercepts of these and plotted it but don't know what it means by finding the solution.
 
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Finding a solution means finding value(s) of x and y that satisfy both equations simultaneously.

Geometrically, it means finding the intersection of the two lines given by the equations you listed.

Why don't you start by choosing one of the equations, and solving for y in terms of x. Then try substituting that into the other equation.
 
y =

2 [x-y=1]
-1 [2x+y=-4]

2x-2y=2
-2x-y=4
--------
-3y=6
y=-2


1[x-y=1]
1[2x+y=-4]

x-y=1
2x+y=-4
--------
-x=-3
x=3 I think I did x wrong somehow but don't know what, but I think I did y right.


like that??
 
homevolend said:
y =

2 [x-y=1]
-1 [2x+y=-4]

2x-2y=2
-2x-y=4
--------
-3y=6
y=-2


1[x-y=1]
1[2x+y=-4]

x-y=1
2x+y=-4
--------
-x=-3
x=3 I think I did x wrong somehow but don't know what, but I think I did y right.


like that??

Everything looked OK until this part:

x-y=1
2x+y=-4
--------
-x=-3

This is wrong because x + 2x does not equal -x.
 
homevolend said:
x-y=1
2x+y= -4

You can add the equations: y will cancel.

ehild
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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