No equations share common solution with this system of equations:
5x+4y=25
2x-7y=26
We have equations
5x+4y=25...(1)
2x-7y=26...(2)
We will solve the equation by elimination method.
Multiply the equation (1) by 2 and equation (2) by 5 and subtracting, we get
10x+8y-(10x-35y)=50-130
43y=-80
[tex]y=\frac{-80}{43}[/tex]
Put [tex]y=\frac{-80}{43}[/tex] in equation (1), we get
[tex]5x+4(\frac{-80}{43} )=25\\x=\frac{279}{43}[/tex]
Now, let us check whether the solution satisfies the other equations
A. 7x+3y=50
[tex]7(\frac{279}{43}) +3(\frac{-80}{43} )\\=\frac{1953-240}{43} \\=39.83\\\neq 50[/tex]
B. 7x-3y=50
[tex]7(\frac{279}{43}) -3(\frac{-80}{43} )\\=\frac{1953+240}{43} \\=\frac{2193}{43} \\=51\\\neq 50[/tex]
C. 5x+4y=2x-7
[tex]5(\frac{279}{43} )+4(\frac{-80}{43} )=2(\frac{279}{43} )-7(\frac{-80}{43} )\\\frac{1395-320}{43} =\frac{558+560}{43} \\\frac{1363}{43} \neq \frac{1118}{43} \\[/tex]
D. 3x-11y=-2
[tex]3(\frac{279}{43}) -11(\frac{-80}{43} )\\=\frac{837+880}{43} \\=\frac{1717}{43} \\=39.9\\ \neq -2[/tex]
E. 3x+11y=-2
[tex]3(\frac{279}{43}) +11(\frac{-80}{43} )\\=\frac{837-880}{43} \\=\frac{-43}{43} \\=-1\\ \neq -2[/tex]
Hence, no equations satisfies the solution with this system of equations:
5x+4y=25
2x-7y=26
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