Two times the width of a certain rectangle exceeds its length by three inches, and eight times its length is twelve more than its perimeter. Find the width of the rectangle.

2x+3=y 8y-12=2(x+y )

exceeds the length, so you'd have to subtract \[2x-3=y\] \[8y-12=2(x+y)\]

Mathmuse has the correct equation. To solve the equation: 8y - 12 = 2(x + y) 8y - 12 = 2x + 2y Use the distributive property 8y -12 = 2x + 2y -2y = - 2y subtract 2y from each side of the equation ------------------ 6y - 12 = 2x 6y - 12 = 2x --- --- --- divide both sides of the equation by 2 to find the value of X 2 2 2 3y - 6 = x now substitute this in the original equation 2 (3y - 6) - 3 = y 6y - 12 - 3 = y Use the distributive property again. 6 y - 15 = y -1 y +15 =-y +15 subtract y from both sides, add 15 to both sides of the equation ----------------- 5 y + 0 = 0y + 15 5y/5 = 15/5 divide both sides of the equation by 5 y = 3 Substitute the value of y in the equation 3y - 6 = x to find the value of x 3 (3) - 6 = x compute 3(3) 3 x 3 = 9 9 - 6 = x compute 9 - 6 which equals 3 3 = x To check: Substitute the values for y (3) and x (3) into the original equations. 8y - 12 = 2(x + y) 8(3) - 12 = 2(3 + 3) 8 x 3 = 24; 3 + 3 = 6 24 - 12 = 2 (6) 24 - 12 = 12; 2 x 6 = 12 12 = 12 The values calculated for x (3) and y (3) are correct. The rectangle is a square having a length and width of three inches.

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