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Find the first of three consecutive integers. The product of the first and the second is 34 less than the square of the third.
Find the first of three consecutive integers. The product of the first and the second is 34 less than the square of the third.
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x=first number
x+1=second number
x+2=third number
x(x+1)=(x+2)^2-34
x^2+x=(x+2)^2-34
x^2+x+34=(x+2)^2
x^2+x+34=x^2+4x+4
0=3x-30
30=3x
x=10.
The first number is 10. Second number is 11. Third number is 12.
x+1=second number
x+2=third number
x(x+1)=(x+2)^2-34
x^2+x=(x+2)^2-34
x^2+x+34=(x+2)^2
x^2+x+34=x^2+4x+4
0=3x-30
30=3x
x=10.
The first number is 10. Second number is 11. Third number is 12.
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x(x+1) = (x+2)^2 - 34
x = 10
x+1 = 11
x+2 = 12
10 times 11 = 12 times 12 - 34
x = 10
x+1 = 11
x+2 = 12
10 times 11 = 12 times 12 - 34