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2x+by=59-2x+by=59\newline 5xy=15x-y=1\newlineIf the system of equations has exactly one solution (x,y)(x,y) and x=2x=2, what is the value of bb?

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Q. 2x+by=59-2x+by=59\newline 5xy=15x-y=1\newlineIf the system of equations has exactly one solution (x,y)(x,y) and x=2x=2, what is the value of bb?
  1. Substitute xx into second equation: Substitute the given value of xx into the second equation.\newlineThe second equation is 5xy=15x - y = 1.\newlineSubstitute x=2x = 2 into this equation.\newline5(2)y=15(2) - y = 1\newline10y=110 - y = 1
  2. Solve for y in second equation: Solve for y in the second equation.\newlineAdd yy to both sides and subtract 11 from both sides to isolate yy.\newline10y+y=1+y10 - y + y = 1 + y\newline10=y+110 = y + 1\newlineSubtract 11 from both sides.\newline101=y10 - 1 = y\newline9=y9 = y
  3. Substitute xx and yy into first equation: Substitute the values of xx and yy into the first equation.\newlineThe first equation is 2x+by=59-2x + by = 59.\newlineSubstitute x=2x = 2 and y=9y = 9 into this equation.\newline2(2)+b(9)=59-2(2) + b(9) = 59\newline4+9b=59-4 + 9b = 59
  4. Solve for b in first equation: Solve for b in the first equation.\newlineAdd 44 to both sides to isolate the term with bb.\newline4+4+9b=59+4-4 + 4 + 9b = 59 + 4\newline9b=639b = 63\newlineDivide both sides by 99 to solve for bb.\newline9b9=639\frac{9b}{9} = \frac{63}{9}\newlineb=7b = 7

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