Practice Makes Perfect Algebra - Carolyn Wheater [12]
y −(− 1) = 3(x − 4)
y + 1 = 3x − 12
y = 3x − 13
To find a line perpendicular to y = 3x − 7 that passes through the point(4, −1), use a slope of .
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EXERCISE 4.11
Determine whether the lines are parallel, perpendicular, or neither.
Find the equation of the line described.
6. Parallel to y = 5x − 3 and passing through the point(3, −1)
7. Perpendicular to 6y − 8x = 15 and passing through the point(−4, 5)
8. Parallel to 4x + 3y = 21 and passing through the point(1, 1)
9. Perpendicular to y = 4x − 3 and passing through the point(4, 13)
10. Parallel to 2y = 4x + 16 and passing through the point(8, 0)
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Systems of linear equations and inequalities
A single equation that contains two or more variables has infinitely many ordered pairs in its solution set. A system of equations is a group of equations, each of which involves two or more variables. For systems with two variables and two equations, the solution is an ordered pair of values that make both of the equations in the system true.
Graphing systems of equations
A system of two linear equations in two variables can be solved by graphing each of the equations and locating the point of intersection. The coordinates of the point of intersection are the values of x and y that solve the system. If the lines are parallel, the system has no solution.
Figure 5.1 Solving a system of equations by graphing.
In Figure 5.1, by graphing 2x − 3y = 6 and y = 3x + 5 on the same set of axes, you can see that they cross at the point(−3, −4). That means that x = −3 and y = −4 will solve both equations. Each equation has an infinite number of solutions, but this one is the only one they have in common.
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EXERCISE 5.1
Graph each system of equations and locate the solution, if one exists.
1. x + y = 6
y = 2x
2. x − y = −6
2x + y = 0
3. x + y = 8
y = 2x − 4
4. x + y = −6
x − 2y = 3
5. 3x + 4y = 12
x + y = 2
6. 2x + y = 3
2x + y = 5
7. x + y = 6
x − y = 8
8. x + y = 7
x − y = 7
9. 2x + 3y = 6
x − y = −7
10. x + 8y = 12
x − 2y = 2
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Graphing systems of inequalities
When you solve a system of equations by graphing, each equation has a graph that is a line, and the solution of the system is the point where the lines intersect. The solution of a system of inequalities is also the intersection of the solution sets of the individual inequalities, but the solution set of an inequality is not just a line, but a region, sometimes called a half-plane. The solution of the system of inequalities is the area where the two half-planes overlap.
To solve the system of inequalities , graph each of the inequalities, shading lightly, then locate the overlap and shade these shared values more prominently(see Figure 5.2).
Figure 5.2 Solving a system of inequalities by graphing.
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EXERCISE 5.2
Graph each system of inequalities and indicate the solution set by shading.
1. y ≥ 2x − 3
y > 3 − x
2. y < x + 5
y ≥ x + 3
3. y ≥ 2x − 3
y ≥ 3 − 2x
4. y > x
y > − x
y < 4
5. y ≤ x + 3
y ≤ 3 − x
y ≥ 0
6. 3x − y > 0
3x − 2y > 4
7. x − y > −5
2x − y < − 2
8. y ≥ 2 x + 2
y < − x + 4
x > − 2
9. y > 2 x + 1
y ≤ 2x − 3
10. y ≥ − 2x − 3
y ≤ x + 2
y ≥ 7x − 3
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Solving systems of equations by substitution
When one equation in a system expresses one variable as a function of the other, like y = 3x − 1 or x = 4 + y, it can be substituted into the other equation. That substitution produces an equation with only one variable, an equation that can be easily solved. To solve the system replace the y in the bottom equation with 3x − 1.
2x + 5y = 12
2x + 5(3x − 1) = 12
2x + 15x − 5 = 12
17x − 5 = 12
17x = 17
x = 1
Once you know the value of one variable, substitute that into one of the original equations to find the value of the other variable.
y = 3x − 1
y = 3(1) − 1
y = 2
Substitution is most convenient when one of the equations is given to you in x = or