Claswork Solve for \( x \) and \( y \) simuteneously, using substitution method a) \( x+y=2 \) and \( 3 x+y=4 \quad x=1 \quad y=1 \) b) \( x+2 y=4 \) and \( x-2 y=0 \quad y=1 \quad x=2 \)

Let's have some fun with these equations! To solve the first set, we start with \( x + y = 2 \). Rearranging gives us \( y = 2 - x \). We then substitute \( y \) into the second equation \( 3x + y = 4 \): \( 3x + (2 - x) = 4 \) This simplifies to: \( 3x + 2 - x = 4 \) Combine like terms: \( 2x + 2 = 4 \) Now, subtract 2 from both sides: \( 2x = 2 \) Dividing by 2, we find \( x = 1 \). Now, substituting back into \( y = 2 - x \): \( y = 2 - 1 = 1 \) So, the solution is \( x = 1 \) and \( y = 1 \)! For the second set of equations, we use \( x + 2y = 4 \). From this, we can express \( x \) as \( x = 4 - 2y \). Next, we substitute this into the second equation \( x - 2y = 0 \): \( (4 - 2y) - 2y = 0 \) Which simplifies to: \( 4 - 4y = 0 \) From here, we can easily solve for \( y \): \( 4y = 4 \) Thus, \( y = 1 \). Plugging this back into \( x = 4 - 2y \): \( x = 4 - 2(1) = 2 \) So, for this pair, we find \( x = 2 \) and \( y = 1 \). Keep those pencils sharp!