### Assignments:

Unfinished Assignment Study Questions for Lesson 30

### Lesson Objectives:

- Solve by graphing.
- Solve by substitution or elimination method.
- Use a system of two linear equations to solve an applied problem.

A system of equations consists of two or more equations considered simultaneously. The corresponding unknowns have the same values. In order to find all the unknown numbers in a system of equations, we must have as many equations as there are unknown numbers.

Solving a system of equations graphically:

Each point at which the graphs intersect is a solution to both equations.

Solve graphically:

x+y = 3 and 2x+y = 0.

Start by solving for y in each equation. So in the first equation, we have y = 3-x, and in the second equation, we have y = -2x.

Now you can plug these equations into your calculator, under y=. And use your calculator to determine the point of intersection. The graphs intersect at a single point, (-3, 6). So (-3, 6) is the solution of the system of equations.

Now let's check our solution. Plug (-3, 6) into our first equation, x+y = 3. So we get (-3)+(6) = 3; 3 = 3, a true statement. Now let's check it with the second equation. So we have 2x+y = 0, or 2(-3)+(6) = 0. This gives us -6+6, or 0=0, another true statement. So (-3, 6) is the solution.

Solving a system of equations algebraically:

Systems of equations are solved by combining the equations so as to obtain a single equation with one unknown number.

And you can accomplish this by substitution or elimination. In substitution, we solve for a variable in one of the equations, and substitute the value of that variable into the second equation.

In elimination, we use addition or subtraction to eliminate one of the variables. And then we solve for the remaining variable and substitute the value for that variable into either of the original equations.

Solve using the substitution method.

x+y = 10 and 2x-3y = -2. We'll call the first equation, 1, and the second equation, 2.

Now let's solve 1 for x. So we get x = 10-y. Then we substitute 10-y in for x in equation 2. So we get 2(10-y)-3y = -2.

So now we have an equation in one variable, which we know how to solve. Let's start by distributing 2 into the parentheses. So we get 20-2y-3y = -2. And then go ahead and combine the -2y and the -3y; so we get -5y. And now if we subtract 20 from both sides, we get -5y = -22. And then divide both sides by -5. So we get y = 22/5.

So now we substitute 22/5 in for y in either of our equations. This is called back substitution. So if we plug it into equation 1, we have x+22/5 = 10. Now subtract 22/5 from both sides, and we have x = 10-22/5. And we can make common denominators by multiplying the top and bottom of 10/1 by 5. So we have 50/5 - 22/5. Now we can just subtract the numerators, 50-22, and get 28/5. So x = 28/5.

So the solution for the system of equations is the ordered pair (28/5, 22/5).

And now we can check our solution by plugging into both of our equations.

So if we plug into 1, we have 28/5+22/5 = 10. This gives us 50/5 = 10, or 10 = 10, a true statement. Now if we plug into equation 2, we have 2(28/5)-3(22/5) = -2. Let's go ahead and divide everything by 2. So we get 28/5-3/2(22/5) = -1. Now 2 goes into 22, 11 times, so we have 28/5 minus 3(11/5), which is 33/5, is equal to -1.

So now we can just subtract the numerators; 28-33 = -5. So we have -5/5 = -1, or -1 = -1, another true statement. So this verifies that (28/5, 22/5) is the solution to our system of equations.

Solve the following system of equations using the elimination method.

x+3y = 8 and x-3y = -5. Since the coefficients of y in both of our equations is 3 and -3, we can eliminate y by adding our equations. So if we go ahead and add them, we're left with 2x = 3, and we get x = 3/2.

Now backsubstitute 3/2 for x in either of our equations. Let's go ahead and plug into our first equation. So we have 3/2+3y = 8. And now if we subtract 3/2 from both sides, we have 3y = 8-3/2. Now 8/1 can be multiplied by 2 in the numerator and denominator so that we get 16/2. So we have 16/2-3/2 = 3y. Now we can subtract the numerators, 16-3 = 13, so we have 13/2 = 3y. And if we multiply both sides by 1/3, the 3's cancel on the left, and we have y = 13/6.

So the solution to our system of equations is the ordered pair, (3/2, 13/6).

There are three types of systems.

The last question we did was called a Consistent Independent System, because there was exactly one solution, or one common point between our two equations. If you were to graph the equations, you would see that we have two perpendicular lines.

Another type of systems is called Inconsistent Independent. Now this is a graph of two parallel lines. There are no common points and no solutions.

And finally, there's a system called Consistem Dependent. These are lines that are identical. There are infinitely many common points, and infinitely many solutions.

A riverboat takes 2 hours to travel downstream 55 kilometers. It takes 3 hours to travel 60 kilometers upstream. What is the speed of the boat and the speed of the current?

We'll need to translate our situation using mathematical language by creating more than one equation with more than one variable. So notice that we're given a time and a distance. Time and distance are related by the following equation: "distance" = "rate"*"time".

Now, downstream, we have that the riverboat takes 2 hours to travel 55 kilometers. And upstream, we have that the boat takes 3 hours to travel 60 kilometers. Now let's go ahead and let the speed of the boat be represented by b, and the speed of the current be represented by c.

Since the boat is going to be traveling with the current, downstream, the rate is b+c. And then since the boat will be traveling against the current, upstream, the rate will be b-c.

So our system of equations is 55 = (b+c)*2 and 60 = (b-c)*3. Now let's go ahead and simplify both of our equations, by dividing by 2 on both sides in our first equation, and dividing by 3 on both sides in our second equation. So we have 27.5 = b+c, and then we have 20 = b-c.

Let's use the elimination method. Notice that we have a positive c, and a -c here. So if we go ahead and add our equations, the c's will cancel, and we're left with 47.5 = 2b, or 23.75 = b, if we divide both sides by 2.

Now let's backsubstitute 23.75 into either one of our equations. I'll use our first equation, 27.5 = b+c. So we have 27.5 = (23.75)+c. And then if we subtract both sides by 23.75, then we're left with c = 3.75.

So the speed of the boat is 23.75 kilometers per hour, and the speed of the current is 3.75 kilometers per hour.