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Solve using elimination.\newline10x+5y=15-10x + 5y = -15\newline3x+5y=20-3x + 5y = 20\newline(_____, _____)

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Q. Solve using elimination.\newline10x+5y=15-10x + 5y = -15\newline3x+5y=20-3x + 5y = 20\newline(_____, _____)
  1. Write Equations: Write down the system of equations to be solved using elimination.\newline10x+5y=15-10x + 5y = -15\newline3x+5y=20-3x + 5y = 20
  2. Eliminate Variable: To use elimination, we want to eliminate one of the variables by adding or subtracting the equations. Since the coefficients of yy are the same in both equations, we can subtract the second equation from the first to eliminate yy.\newlineSubtract the second equation from the first:\newline(10x+5y)(3x+5y)=1520(–10x + 5y) - (–3x + 5y) = –15 - 20
  3. Solve for x: Perform the subtraction to eliminate yy and solve for xx.
    10x+5y(3x)5y=1520-10x + 5y - (-3x) - 5y = -15 - 20
    10x+3x=35-10x + 3x = -35
    7x=35-7x = -35
  4. Substitute x: Divide both sides by 7–7 to solve for x.\newlinex=357x = \frac{–35}{–7}\newlinex=5x = 5
  5. Solve for y: Now that we have the value of xx, we can substitute it back into one of the original equations to solve for yy. Let's use the first equation:\newline10x+5y=15–10x + 5y = –15\newlineSubstitute x=5x = 5 into the equation:\newline10(5)+5y=15–10(5) + 5y = –15
  6. Final Solution: Perform the multiplication and solve for yy.50+5y=15-50 + 5y = -155y=15+505y = -15 + 505y=355y = 35
  7. Final Solution: Perform the multiplication and solve for yy.50+5y=15-50 + 5y = -155y=15+505y = -15 + 505y=355y = 35Divide both sides by 55 to solve for yy.y=355y = \frac{35}{5}y=7y = 7

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