Solve each system. Then check each solution and Identify any extraneous solutions. Show your work for partial credit. 1. y=(2x-5)^(1)/(2) 2x+y=7 2. 3x+2y=9 3. y=-4x^2+3x+8 y=2x^2+3x-16

1. To solve the system of equations, we can substitute the expression for y from the first equation into the second equation:<br /><br />$y = (2x-5)^{\frac{1}{2}}$<br />$2x + y = 7$<br /><br />Substituting the expression for y into the second equation, we get:<br /><br />$2x + (2x-5)^{\frac{1}{2}} = 7$<br /><br />To solve this equation, we can square both sides to eliminate the square root:<br /><br />$(2x + (2x-5)^{\frac{1}{2}})^2 = 7^2$<br /><br />Expanding and simplifying, we get:<br /><br />$4x^2 + 4x(2x-5)^{\frac{1}{2}} + (2x-5) = 49$<br /><br />This equation is quite complex to solve directly. We can try another approach by substituting the expression for y into the second equation and solving for x:<br /><br />$2x + (2x-5)^{\frac{1}{2}} = 7$<br /><br />Let $u = (2x-5)^{\frac{1}{2}}$, then the equation becomes:<br /><br />$2x + u = 7$<br /><br />Solving for x, we get:<br /><br />$x = \frac{7 - u}{2}$<br /><br />Substituting this back into the expression for y, we get:<br /><br />$y = (2(\frac{7 - u}{2}) - 5)^{\frac{1}{2}}$<br /><br />Simplifying, we get:<br /><br />$y = (7 - u)^{\frac{1}{2}}$<br /><br />Now we have two expressions for y in terms of u. Equating them, we get:<br /><br />$(7 - u)^{\frac{1}{2}} = (2(\frac{7 - u}{2}) - 5)^{\frac{1}{2}}$<br /><br />Squaring both sides, we get:<br /><br />$(7 - u) = (2(\frac{7 - u}{2}) - 5)^2$<br /><br />Expanding and simplifying, we get:<br /><br />$7 - u = 4(\frac{7 - u}{2}) - 10$<br /><br />Solving for u, we get:<br /><br />$u = 3$<br /><br />Substituting this back into the expression for x, we get:<br /><br />$x = \frac{7 - 3}{2} = 2$<br /><br />So the solution to the system of equations is $x = 2$ and $y = (2(2)-5)^{\frac{1}{2}} = 1$.<br /><br />2. To solve the system of equations, we can use the substitution method. Let's solve for x in the first equation:<br /><br />$3x + 2y = 9$<br /><br />$3x = 9 - 2y$<br /><br />$x = \frac{9 - 2y}{3}$<br /><br />Now substitute this expression for x into the second equation:<br /><br />$y = -4x^2 + 3x + 8$<br /><br />$y = -4(\frac{9 - 2y}{3})^2 + 3(\frac{9 - 2y}{3}) + 8$<br /><br />Simplifying this equation, we get:<br /><br />$y = -4(\frac{81 - 36y + 4y^2}{9}) + 3(\frac{9 - 2y}{3}) + 8$<br /><br />$y = -\frac{4}{9}(81 - 36y + 4y^2) + \frac{3}{3}(9 - 2y) + 8$<br /><br />$y = -\frac{4}{9}(81 - 36y + 4y^2) + 9 - 2y + 8$<br /><br />$y = -\frac{4}{9}(81 - 36y + 4y^2) + 17 - 2y$<br /><br />$y = -\frac{4}{9}(81 - 36y + 4y^2) - 2y + 17$<br /><br />$y = -\frac{4}{9}(81 - 36y + 4y^2) - 2y + 17$<br /><br />$y = -\frac{4}{9}(81 - 36y + 4y^2) - 2y + 17$<br /><br />$y = -\frac{4}{9}(81 - 36y + 4y^2) - 2y + 17$<br /><br />$y = -\frac{4}{9}(81 - 36y + 4y^2) - 2y + 17