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For each order paid determine whether it is a solution to 5x+7y=-3

For Each Order Paid Determine Whether It Is A Solution To 5x7y3 class=

Sagot :

Given equation is

[tex]5x+7y=-3[/tex]

Take the first point (5,-4), and substitute 5 for x and -4 for y in the left hand side of the equation:

[tex]\begin{gathered} 5(5)+7(-4)=25-28 \\ =-3 \end{gathered}[/tex]

So, the left hand side value -3 is equal to the right hand side value -3.

Therefore, the point (5,-4) is a solution to 5x+7y=-3.

Take the point (-2,1). Substitute -2 for x and 1 for y in the left hand side of the equation:

[tex]\begin{gathered} 5(-2)+7(1)=-10+7 \\ =-3 \end{gathered}[/tex]

Here, also the left hand side value -3 is equal to the right hand side value -3.

Therefore, the point (-2,1) is a solution to 5x+7y=-3.

Next take the point (6,0) and substitute 6 for x and 0 for y in left hand side of the equation:

[tex]\begin{gathered} 5(6)+7(0)=30+0 \\ =30 \end{gathered}[/tex]

Notice that the left hand side value 30 is not equal to right side value -3

So, the point (6,0) is not a solution to the equation 5x+7y=-3.

For (-4,-3) substitute -4 for x and -3 for y in left hand side of the equation:

[tex]\begin{gathered} 5(-4)+7(-3)=-20-21 \\ =-41 \end{gathered}[/tex]

Notice that the left hand side value -41 is not equal to right side value -3

So, the point (-4,-3) is not a solution to the equation 5x+7y=-3.

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