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How many solutions does the following equation have?

7(y+3)=5y+8
Choose 1 answer:
(A) No solutions
(B) Exactly one solution
(c) Infinitely many solutions

How many solutions does the following equation have?\newline7(y+3)=5y+87(y+3)=5y+8\newlineChoose 11 answer:\newline(A) No solutions\newline(B) Exactly one solution\newline(C) Infinitely many solutions

Full solution

Q. How many solutions does the following equation have?\newline7(y+3)=5y+87(y+3)=5y+8\newlineChoose 11 answer:\newline(A) No solutions\newline(B) Exactly one solution\newline(C) Infinitely many solutions
  1. Distribute the 77: Distribute the 77 to both terms inside the parentheses.\newline7(y+3)=7y+737(y+3) = 7\cdot y + 7\cdot 3\newline=7y+21= 7y + 21
  2. Rewrite the equation: Rewrite the equation with the distributed terms. 7y+21=5y+87y + 21 = 5y + 8
  3. Move terms with yy: Move all terms containing yy to one side of the equation by subtracting 5y5y from both sides.\newline7y+215y=5y+85y7y + 21 - 5y = 5y + 8 - 5y\newline=2y+21=8= 2y + 21 = 8
  4. Move constant term: Move the constant term on the y-side to the other side by subtracting 2121 from both sides.\newline2y+2121=8212y + 21 - 21 = 8 - 21\newline=2y=13= 2y = -13
  5. Solve for y: Solve for y by dividing both sides by 22.\newline2y2=132\frac{2y}{2} = \frac{-13}{2}\newliney=6.5\Rightarrow y = -6.5

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