Algebra 1 dba 5linear equation: an equation that looks like a line on a graph. the slope is constant. system of equations: a set of equations (at least 2) that are graphed on the coordinate plane. point of intersection: this is the point the lines on the graph, and it is the solution of the system of equations. graph the 2 lines.    if the lines are parallel, the slopes are the same and the y-intercepts are different. there are solutions. if the lines are the same line, the slopes are the and the y-intercepts are the same. there have infinitely many solutions. if the lines are intersecting, the slopes are different and the y intercepts can be the same. there is solution. substitution method – solve for one variable in one equation and then substitute into the other equation. example:     x + 4y = 3                     2x – 3y = 17 solve for x in the first equation:   x = -4y + 3 now substitute -4y + 3 in place of the x in the second equation. 2(-4y + 3) – 3y = 17   distribute the 2 over the parenthesis. -8y + 6 – 3y = 17   combine like terms. -11y + 6 = 17  minus 6 from both sides. -11y = 11  divide both sides by -11. y = -1 but wait!  you still need to find the value of the other variable. so plug -1 into either equation.  the first equation will be easiest. try it: … . … . x = so the answer is ( ) which is the point of intersection. elimination method – get one variable to have the same coefficient but opposite signs so they will cancel each other out when the 2 equations are added. example:      2x – 5y = -3                      -x + 2y = 3 multiply the entire second equation by 2: 2(-x + 2y) = 2(3) simplify: … . … . add the x’s, the y’s, and the numbers (the x’s should cancel each other out) we get: -y = 3  divide both sides by -1 (you can’t have a negative variable) y = -3 but wait!  you still need to find the value of the other variable. plug -3 in place of y. you can choose either equation.  i chose the 2nd equation. -x + 2(-3) = 3 -x = 9 x = -9 so the answer is which is the point of intersection. linear inequalities will usually have x and y variables to graph them – you have to convert to y=mx+ b: example: 3x – 2y < = 4 minus 3x on both sides:  -2y < = -3x + 4 divide all terms by -2:   y > = -3/2x + 2    notice that i flipped the inequality symbol. now you can determine how the graph will look. if the inequality is < or > , draw a __dashed _ line. if the inequality is < = or > =, draw a line. if the inequality is < or < =, shade __below (less than… the line. if the inequality is > or > =, shade the line steps for solving systems of inequalities by graphing ·graph both inequalities on the same coordinate plane. the line should be solid if the inequality is __ _or and dashed if the inequality is or we shade above the line if the inequality is or we shade below the line if the inequality is or ·emphasize the area where the shading is overlapped. test an ordered pair from the overlapping shaded area check that both inequalities in the system are true. for example: y> =2/3x-4 and y> 2x-5