sketch the graph of the set of all points that solve each system of linear inequalitiesSo this is algebra 1 review for algebra 2 and i havent done it in forever, step by step explanation pls for both?​

Accepted Solution

Explanation:For our purposes here, we will call these relations "equations" simply because it is less work than calling them "relations" or "inequalities" in the cases where an equal sign is not involved.22. These equations do two things:they put restrictions on the values that x and y may take onthey put restrictions on the relationships between x and y.It is convenient to rewrite the equations to identify these separate effects.   0 < x ≤ 125 . . . . . a restriction on the possible values of x   x ≥ 2y > 0 . . . . . . this can be interpreted 3 ways:      x > 0 . . . . already covered by the first equation      y > 0 . . . . what you get when you divide by 2. This is a restriction on possible values of y      x ≥ 2y . . . . a restriction on the relationship between x and y. This can be rewritten as ...      y ≤ (1/2)x . . . . divide by 2   2x +2y ≤ 300 . . . . . a restriction on the relationship between x and y, more conveniently written as ...      x + y ≤ 150__In each of these "restrictions on relationship", the boundary (equal to case) is an equation for a straight line. Those lines are ...y = (1/2)xx+y = 150The first goes through the point (0, 0) and has a slope of 1/2, meaning it rises 1/2 unit for each 1 unit of "run" to the right. The first attachment shows a graph scaled so that it is easy to see a rise of 50 for a run of 100. Note that we don't plot anything for x-values greater than 125, which is the restriction the first original equation puts on values of x. Since y may be less than (1/2)x, the graph is shaded below the line (and above y=0, the limit on the possible values of y). y=0 is not included in the graph, so that boundary line can be dashed.The second has x- and y-intercepts of 150. The line would extend between the points (0, 150) and (150, 0) but for the fact that no values of x above 125 are allowed. Once again, x and y may be less than the values on this boundary line, so the shading is below (and to the left) of the line, and only in the first quadrant. Neither x=0, or y=0 are part of the solution set.The set of all points that solve the system of equations is the doubly-shaded area. __23. Here, the inequalities that identify x- and y- values of interest are already separated:x > 0y > 0These tell you the solution set is restricted to the first quadrant, and x=0 or y=0 are not part of the solution.The boundary equations have no equal sign, so when you draw the corresponding lines, you make those lines dashed.It can be useful to consider the x- and y-intercepts of each of the equations ...2x + y = 10x - 2y = 8You can find these by setting the opposite variable to zero and dividing by the coefficient of the variable of interest.For example, the x-intercept for the first equation is found by solving ...   2x + 0 = 10   x = 10/2 = 5And the y-intercept is found by solving ...   0 + y = 10   y = 10The (dashed) boundary line will be the line between these points (5, 0) and (0, 10).For the second equation, similar work gives intercept points of (8, 0) and (0, -4). Since we're only interested in the part of this line that's in the first quadrant, we can use the point (0, -4) to help us draw the graph, but that point and the line below y=0 will not be part of the solution set. Once again, the line will be dashed.Both x and y have positive coefficients in the first equation, and the relation is "less than" so shading is below the line.x has a positive coefficient in the second equation, and that relation is also "less than", so shading is to the left of the line. The y-coefficient is negative, so y-values that are in the solution set will be above, not below, the boundary line (same as to the left of it).See the second attachment for a graph.The set of all points that solve the system of equations is the doubly-shaded area. (You will note that this is the same as the shaded area for the first inequality. The second inequality adds no restrictions to the solution space.)