Maintaining Mathematical Proficiency

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Exercises 1 To identify the (x,y) ordered pair to which a given point corresponds, count horizontally from the point to the y-axis to find the x-coordinate. Similarly, count vertically from the point to the x-axis to find the y-coordinate.In this case, we have identified that the point G lies at the coordinate pair (-5,-2).
Exercises 2 To identify the (x,y) ordered pair to which a given point corresponds, count horizontally from the point to the y-axis to find the x-coordinate. Similarly, count vertically from the point to the x-axis to find the y-coordinate.In this case, we have identified that the point D lies at the coordinate pair (2,0).
Exercises 3 Quadrant I is the region that contains points having both positive x- and y-coordinates. Only point C(3,5) lies within this region.Since 0 is neither positive nor negative, points B(0,4) and D(2,0) are not in any quadrant, they lie on the axes.
Exercises 4 Quadrant IV is the region that contains points with positive x-coordinates and negative y-coordinates. Only point E(3,-3) lies within this region.Since 0 is neither positive nor negative, point D(2,0) is not in any quadrant because it lies on the axis.
Exercises 5 To solve the given literal equation for y, we will use the Properties of Equality to apply inverse operations on the equation. Remember, only like terms can be combined. x−y=5LHS−x=RHS−x-y=-x+5LHS/-1=RHS/-1-1-y​=-1-x​+-15​Simplify termsy=x−5
Exercises 6 To solve the given literal equation for y, we will use the Properties of Equality to apply inverse operations on the equation. Remember, only like terms can be combined. 6x+3y=-1LHS−6x=RHS−6x3y=-6x−1LHS/3=RHS/333y​=3-6x​−31​Simplify termsy=-2x−31​
Exercises 7 To solve the given literal equation for y, we will use the Properties of Equality to apply inverse operations on the equation. Remember, only like terms can be combined. 0=2y−8x+10LHS+8x=RHS+8x8x=2y+10LHS−10=RHS−108x−10=2yLHS/2=RHS/228x​−210​=22y​Simplify terms4x−5=yRearrange equationy=4x−5
Exercises 8 To solve the given literal equation for y, we will use the Properties of Equality to apply inverse operations on the equation. Remember, only like terms can be combined. -x+4y−28=0LHS+28=RHS+28-x+4y=28LHS+x=RHS+x4y=x+28LHS/4=RHS/444y​=4x​+428​Simplify termsy=41​x+7
Exercises 9 To solve the given literal equation for y, we will use the Properties of Equality to apply inverse operations on the equation. Remember, only like terms can be combined. 2y+1−x=7xLHS+x=RHS+x2y+1=8xLHS−1=RHS−12y=8x−1LHS/2=RHS/222y​=28x​−21​Simplify termsy=4x−21​
Exercises 10 To solve the given literal equation for y, we will use the Properties of Equality to apply inverse operations on the equation. Remember, only like terms can be combined. y−4=3x+5yLHS+4=RHS+4y=3x+5y+4LHS−5y=RHS−5y-4y=3x+4LHS/-4=RHS/-4-4-4y​=-43x​+-44​Simplify termsy=-43​x−1
Exercises 11 Let's check each quadrant one by one.Quadrant I In Quadrant I, both x and y are positive numbers. Let's choose (2,2) as our point. Then we multiply both coordinates by a negative number, let's use -1. This reverses the point's signs so we get (-2,-2). Let's plot these points.Quadrant II In Quadrant II, x is negative and y is positive. Let's choose (-1,1) as our point. Then we multiply both coordinates by a negative number, let's use -1. This reverses the point's signs so we get (1,-1). Let's plot these points.Quadrant III In Quadrant III, both x and y are negative numbers. Let's choose (-3,-3) as our point. Then we multiply both coordinates by a negative number, let's use -1. This reverses the point's signs so we get (3,3). Let's plot these points.Quadrant IV In Quadrant IV, x is positive and y is negative. Let's choose (4,-4) as our point. Then we multiply both coordinates by a negative number, let's use -1. This gives the point (-4,4). Let's plot these points.Conclusion In each case, multiplying the point by a negative number carries the point diagonally across the origin to the new point. This is called a reflection about the origin.