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-28 The correct answer is . A system of two linear equations in two variables, x and y, has no solution if the lines represented by the equations in the xy-plane are distinct and parallel. The graphs of two lines in the xy-plane represented by + = equations in the form Ax By C, where A, B, and C are constants, are parallel if the coefficients for x and y in one equation are proportional to the corresponding coefficients for x and y in the other equation. The first equation in 48 -64 =48 +24 + = the given system, x y y , can be written in the form Ax By C by 48 48 -112 =24 subtracting y from both sides of the equation to yield x y . The = 1 -12 second equation in the given system, ry 8 x, can be written in the form + = 12 12 + = 1 Ax By C by adding x to both sides of the equation to yield x ry 8. The 1 coefficient of x in the second equation is 4 times the coefficient of x in the first 48 1 =12 equation. That is, b 4 l . For the lines to be parallel, the coefficient of y in the 1 second equation must also be 4 times the coefficient of y in the first equation. -112 1 = -28= Therefore, b 4 l r, or r. Thus, if the given system has no solution, the -28 value of r is . 52 SAT PRACTICE TEST #6 ANSWER EXPLANATIONS