Solve for X and Y: 2y + x = 7; 3y - x = 8

x=1, y=3; Firstly, rearranging for x in the first equation: x = (7-2y); Substitute this into the second equation: 3y - (7 -2y) = 8; Expand: 3y -7 +2y = 8 (watch negative) 3y +2y = 15 Therefore, 5y = 15 and y=3. Now substitute Y into either of the equations to find x: 2(3) + x = 7 6 + x = 7 Therefore x = 1.

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Answered by Ruth K. Maths tutor

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Jo wants to work out the solutions of x^2 + 3x – 5 = 0 She says, ‘‘The solutions cannot be worked out because x^2 + 3x – 5 does not factorise to (x + a)(x + b) where a and b are integers.’’ Is Jo correct?


Expand and simplify: 5(x + 3) - 3(y - 2)


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