Solving Systems Algebraically

Solving Systems Algebraically

Solving Systems Algebraically

Memory Lane: Solving a System Algebraically

To solve a system of equations algebraically, you can use the method of Substitution or Elimination.

Solving by Substitution

1. Label your equations as ① and ②.
2. Isolate a variable from one of the equations. Choose the easiest one.
3. Substitute the expression for this variable into the other equation. If you isolate from ①, substitute into ②, and vice versa. (Hint: You should now only have the variable, y.)
4. Solve for the remaining variable.
5. Substitute the known value of the variable you just solved into one of the equations to solve for the other variable.
6. Verify your solution and state it as an ordered pair (x, y).

Solving by Elimination

1. Label your equations as ① and ②.
2. Eliminate a variable by adding or subtracting the like terms of each equation together. (For this to happen, the variable you are eliminating must have the same coefficient. You will often eliminate y.)
3. Solve for the variable that is not eliminated.
4. Substitute the known value of the variable you just solved into one of the equations to solve for the other variable.
5. Verify your solution and state it as an ordered pair (x, y).

Class Example 1

Solve the following Linear System using the Substitution Method:

\begin{cases}
2x + y = 5 &①\
x + y = 3 &②
\end{cases}

Isolate y in equation ②:

y = 3 - x

Substitute y = 3 - x into equation ①:

2x + (3 - x) = 5

2x + 3 - x = 5

x = 2

Now, solve for y by substituting x = 2 into either equation ① or ②. Using equation ②:

2 + y = 3

y = 1

Therefore, the solution is (2, 1).

Class Example 2

Solve the following Linear-Quadratic System using the Substitution Method:

\begin{cases}
5x - y = 10 &①\
x^2 + x - 2y = 0 &②
\end{cases}

1. Label each equation.

2. Isolate a variable from one of the equations (always choose y). From equation ①:

   y = 5x - 10

3. Substitute the expression into equation ②:

   x^2 + x - 2(5x - 10) = 0

4. Solve for the variable:

   
   x^2 + x - 10x + 20 = 0
   

   
   x^2 - 9x + 20 = 0
   

   
   (x - 5)(x - 4) = 0
   

   
   x = 4 \text{ or } x = 5
   

5. Solve for the other variable:

   • When x = 4:
     
     5(4) - y = 10
     

     
     y = 10
     

   • When x = 5:
     
     5(5) - y = 10
     

     
     y = 15
     

6. Verify the solutions:

   • Verify the solution (4, 10):
     • LS (Left Side): 5x - y = 5(4) - 10 = 20 - 10 = 10
     • RS (Right Side): 10
     • Thus, 10 = 10 \checkmark
       *Verify the solution (4,10):
     • LS (Left Side): x^2 + x - 2y = (4)^2 + (4) - 2(10) = 16 + 4 - 20 = 0
     • RS (Right Side): 0
     • Thus, 0 = 0 \checkmark

Class Example 3

Determine two integers where the sum of the smaller number and twice the larger number is 46. Also, when the square of the smaller number is decreased by three times the larger number, the result is 93.

a. Write a Linear-Quadratic System to represent the context above:

Let:
*   x be the smaller integer
*   y be the larger integer

The system is:


\begin{cases}
x + 2y = 46 &①\\
x^2 - 3y = 93 &②
\end{cases}

b. Solve the system using the Elimination Method:

To eliminate y, we need to make the coefficients of y equal.  Multiply equation ① by 3 and equation ② by 2:


\begin{cases}
3(x + 2y) = 3(46) \rightarrow 3x + 6y = 138 &①\\
2(x^2 - 3y) = 2(93) \rightarrow 2x^2 - 6y = 186 &②
\end{cases}


Add the equations to eliminate y:


(0x^2 + 3x + 6y) + (2x^2 + 0x - 6y) = 138 + 186



2x^2 + 3x = 324


2x^2 + 3x - 324 = 0

Solve the quadratic equation using the quadratic formula:

For ax^2 + bx + c = 0, x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Here, a = 2, b = 3, c = -324

x = \frac{-3 \pm \sqrt{(3)^2 - 4(2)(-324)}}{2(2)}

x = \frac{-3 \pm \sqrt{9 + 2592}}{4}

x = \frac{-3 \pm \sqrt{2601}}{4}

x = \frac{-3 \pm 51}{4}

x = 12 \text{ OR } x = -13.5

Since we are looking for integers, we discard x = -13.5.

So, x = 12.  Substitute this into equation ①:


12 + 2y = 46



2y = 34



y = 17


Therefore, the numbers are 12 and 17.

Class Example 4

A Canadian cargo plane drops a crate of emergency supplies to aid-workers on the ground. The crate is in free fall until its parachute opens to bring it gently to the ground. The crate's height, h, in meters, above the ground t seconds after leaving the aircraft is given by the following equations:

\begin{cases}
h = -4.9t^2 + 700 & \text{models height during free fall} \
h = -5t + 650 & \text{models height after parachute opens}
\end{cases}

Notice the variables are height and time.

a. How long after the crate leaves the aircraft does the parachute open? Write your answer to the nearest hundredth of a second.

We need to find where the height functions intersect. At the point of intersection:


-4.9t^2 + 700 = -5t + 650



0 = -4.9t^2 + 5t - 50


Using the quadratic formula, where a = -4.9, b = 5, and c = 50:


t = \frac{-5 \pm \sqrt{(5)^2 - 4(-4.9)(50)}}{2(-4.9)}



t = \frac{-5 \pm \sqrt{25 + 980}}{-9.8}



t = \frac{-5 \pm \sqrt{1005}}{-9.8}

t = 3.75 \text{ OR } t = -2.73

Since time cannot be negative, t = 3.75 seconds

b. How high above the ground is the crate when the parachute opens? Write your answer to the nearest meter.

Substitute t = 3.75 into equation ②:

h = -5(3.75) + 650

h = -18.75 + 650

h = 631.25 \approx 631 \text{ meters}

Class Example 5

Solve the following system of equations algebraically. Verify your solution(s).

\begin{cases}
3x^2 - x - y - 2 = 0 &①\
6x^2 + 4x - y - 4 = 0 &②
\end{cases}

Subtract equation ① from equation ②:

(6x^2 + 4x - y - 4) - (3x^2 - x - y - 2) = 0


3x^2 + 5x - 2 = 0


3x^2 + 6x - x - 2= 0


3x(x+2) -1(x+2) = 0


(3x-1)(x+2)=0

So,
x = \frac{1}{3}, or x = -2
Solve for y when x = \frac{1}{3} using equation ①

3(\frac{1}{3})^2 - \frac{1}{3} - y - 2 = 0


\frac{1}{3} - \frac{1}{3} - y -2 = 0

So, y = -2. One solution is ( \frac{1}{3}, -2)
Now solve for y when x = -2 using equation ①

3(-2)^2 - (-2) - y - 2 = 0


12 + 2 - y - 2 = 0

So, y = 12. Another solution is (-2, 12)

Class Example 6

A 250-g ball is thrown into the air with an initial velocity of 22.36 m/s. The kinetic energy, Ek, of the ball is given by Ek = \frac{5}{32}(d-20)^2
and its potential energy, Ep, is given by Ep = \frac{5}{32}(d-20)^2 + 62.5
where energy is measured in joules (J) and d is the horizontal distance traveled by the ball, in meters.

a. At what distance does the ball have the same amount of kinetic energy as potential energy?
When Ek = Ep, we have:
\frac{5}{32}(d-20)^2 = -\frac{5}{32}(d-20)^2 + 62.5
Since Ek = Ep,
\frac{5}{32}(d-20)^2 = -\frac{5}{32}(d-20)^2 + 62.5
After multiplying both sides by 32 and then dividing by 5, the equation changes to:
(\frac{32}{5}) * \frac{5}{32}(d-20)^2 = (\frac{32}{5}) * [-\frac{5}{32}(d-20)^2 + 62.5]
\rightarrow 2(d-20)^2 = 400
\rightarrow (d-20)^2 = 200
\rightarrow d-20 = \pm \sqrt{200}
\rightarrow d = 20 \pm \sqrt{200}
\rightarrow d = 20 \pm 10\sqrt{2}
So, d = 34.14 m or d = 5.86 m

b. How many joules of each type of energy does the ball have at these distances?
Potential energy is also 31.25 Joules
When d = 5.86 m,
Ek = \frac{5}{32}(5.86-20)^2 Ek = \frac{5}{32}(-14.14)^2
Ek = \frac{5}{32}(199.94) Ek = 31.24
Therefore, the Kinetic energy Ek is 31.25 Joules and the Potential energy Ep is 31.25 Joules

Key Ideas

• To solve a system of equations means to find the point(s) of intersection, (x, y).
• To solve a system of equations algebraically, use the method of Substitution or Elimination.
• For Linear-Quadratic or Quadratic-Quadratic Systems, always isolate for or eliminate the variable, y since you must be left with only one variable.
• Any system of equations can be solved using the graphing calculator functions.