Solving Systems of Equations with Absolute Value, Polynomials, and Inequalities


 Lionel Brooks
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1 Solving Systems of Equations with Absolute Value, Polynomials, and Inequalities Solving systems of equations with inequalities When solving systems of linear equations, we are looking for the ordered pair where two or more lines intersect. But when solving systems of linear inequalities, we are now looking for the areas where two inequalities intersect. Find a solution to the system of inequalities y > 3x + 5 and y x. We can solve this system by graphing. Notice that the strict inequality in y > 3x + 5 is dotted to demonstrate that a solution does not fall along that line. The area above the dotted line is shaded to show that any ordered pair above the line will satisfy the inequality. Similarly, any ordered pair in the area below the line y x will satisfy the second inequality. The purple region is where both areas overlap. Any ordered pair in this area will be a solution to the system. Find two solutions and two nonsolutions to the following system of equations: 8x + 4y 1 { y > 1 x We can solve this using substitution, just like a normal system of equations. Why? We re looking for the boundaries of the region that solves the system, which will be defined by the lines which would solve this system if we had equal signs rather than inequalities. 8x + 4 ( 1 x ) 1
2 8x + x x x 0 x We then plug our value for x back into one of the original equations to solve for y: y > 1 x y > 1 () y > 1 y > 1 We have discovered the ordered pair (, 1) where the two lines intersect. But (, 1) is not a solution we know, for example, that y must be greater than 1. Let s take a look at what this looks like on a graph: Two solutions to this equation might be (1, 0) or (,3). Two nonsolutions might be (0,5) or (3,0). Find the solution set for the following system of inequalities: x 5y 6 { x 5y > If we solve both for y, we get the following two linear inequalities:
3 y { 5 x y < 5 x We should notice first that these will be parallel lines. However, when we graph them we should also notice that the shaded regions will not intersect one is above the yintercept 6/5 and one is below the yintercept . So there is no solution to this system of inequalities. Solving systems of equations with polynomials Find all solutions that satisfy the following system of equations: y = 5x + and y = x 4. Notice that the first equation is a firstdegree linear equation, while the second is a seconddegree quadratic. This means that there may be 0, 1, or points of intersection. We can solve systems of equations with polynomials using the same methods we learned earlier. This system is a good candidate for using substitution: y = 5x + y = x 4 5x + = x 4 After we substitute, we can solve this polynomial the same way we would normally by factoring or using the quadratic equation. 5x + = x x = x  65x 5x 0 = x 5x 6 0 = (x 6)(x + 1) x= 6 or 1 From here we can plug our solution back into the original equation to solve for y. y = 5(6) + y = 3 y = 5(1) + y = 3
4 So the two solutions are (6, 3) and (1, 3). We can check these by plugging them into the second equation as well. Here are the two solutions represented graphically. Solve the following system: We can begin by substituting in for y: { y = x x 6 y = x 10 x 10 = x x 6 10 = x x 6 x x + 4 = 0 x = ( ) ± ( ) 4(1)(4) (1) ± 4 16 x = x = ± 1 We could continue simplifying from here, but look what happened we will need to take the square root of a negative number and our roots will include imaginary numbers. What s happening? Looking at the graph makes it clear these two functions never intersect. We found imaginary roots because there is no solution to this system.
5 Solving systems of equations with absolute value Solve the system of equations y = 3x + 1 and y = x 4x + 1. We know that the first absolute value function will be shaped like a v and the second function will be a parabola. Therefore, there will generally be  4 points of intersection (unless the two functions share a vertex). The absolute value parabola can be considered a piecewise function with two linear equations. In order to plot this graph, we can begin by finding the vertex where y = 0: 3x + 1 = x = 1 /3 /3 x = 1 3 Since this is an upwardfacing parabola, we know that from x ( 1, ), the linear equation will be 3 y = 3x + 1. From x (, 1 ), the linear equation will be: 3 y = 3x + 1 /1 /1 y = 3x 1 Now we can solve the following system of inequalities the same way we would solve a normal system of equations with polynomials: 3x 1 if x (, 1 y = { 3 ) 3x + 1 if x ( 1 3, ) y = x 4x + 1
6 We can use substitution to solve, beginning with the first equation in our piecewise function: y = 3x 1 y = x 4x + 13x 1 = x 4x x + 3x 1 = x x = x x + From here, we will need to use the quadratic equation in order to solve: x = ( 1) ± ( 1) 4()() () 1 ± 1 16 x = 4 x = 1 ± 15 4 Our solution is imaginary number! This means these two equations do not intersect. If they had, we would also need to ensure our solution fell within the function s domain. Let s solve this system of equations for the second part of the piecewise equation. y = 3x + 1 y = x 4x + 1 3x + 1 = x 4x + 13x  3x 1 = x 7x = x 7x 0 = x (x 7) x = 0 or 7 We can now plug our xvalues in to solve for y: y = 3(0) + 1 y = 1
7 y = 3( 7 ) + 1 y = 1 + y = 3 We have two solutions to our equation: (0, 1) and ( 7, 3 ). Here is the graph of the solutions:
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