Will someone please help me

Answer:
The filled table for each equation by using the exact values in the table is
10x+2y=56
x y
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0 28
[tex]\frac{56}{10}[/tex] 0
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8x+3y=49
x y
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0 [tex]\frac{49}{3}[/tex]
[tex]\frac{49}{8}[/tex] 0
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Step-by-step explanation:
Given equations are 10x+2y=56 and 8x+3y=49
To fill the table for each equation by using the exact values in the table :
10x+2y=56
put x=0 in above equation we get
10(0)+2y=56
2y=56
[tex]y=\frac{56}{2}[/tex]
[tex]y=28[/tex]
Therefore (0,28)
put y=0 in the given equation 10x+2y=56 we get
10x+2(0)=56
10x=56
[tex]x=\frac{56}{10}[/tex]
Therefore ([tex]\frac{56}{10}[/tex],0)
10x+2y=56
x y
_________________________
0 28
[tex]\frac{56}{10}[/tex] 0
__________________________
For
8x+3y=49
put x=0 in above equation we get
8(0)+3y=49
3y=49
[tex]y=\frac{49}{3}[/tex]
Therefore (0,[tex]\frac{49}{3}[/tex])
put y=0 in the given equation 8x+3y=49 we get
8x+3(0)=49
8x=49
[tex]x=\frac{49}{8}[/tex]
Therefore ([tex]\frac{49}{8}[/tex],0)
8x+3y=49
x y
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0 [tex]\frac{49}{3}[/tex]
[tex]\frac{49}{8}[/tex] 0
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