I’m not sure where exactly you mean “final” in the solving process. So, I’ve prepared a couple of problems that I will through slowly and carefully, showing all the steps to the final answer.

Example 1: Solve the following system of equations-##2x + y = 5, 3x + 2y = 9##

Since we want to solve with substitution, we must solve for one variable in one of the equations. I think it would be easiest to solve for ##y## in the first equation.

##y = 5 – 2x##

We can now substitute into the other equation:

##3x + 2(5 – 2x) = 9##

##3x + 10 – 4x = 9##

##-x = -1##

##x = 1##

We must now find the value of ##y##. This is found by inserting ##x = 1## into one of the equations and solving for ##y##.

##y = 5 – 2x##

##y = 5 – 2(1)##

##y = 3##

Hence, our solution set is ##{1, 3}##.

Example 2: Find all real values of ##x## and ##y## that satisfy the following system of equations: ##3y = -2x^2 + 2, 2x^2 – 3y^2 = -4##

Once again, as with the last example, we need to solve for one of the variables in one of the equations. It looks easiest to isolate the ##y## in the first equation, however solving this equation won’t be as neat as solving the previous one.

##y = -2/3x^2 + 2/3##

We can now substitute into equation 2.

##2x^2 – 3(-2/3x^2 + 2/3)^2 = -4##

##2x^2 – 3(4/9x^4 – 8/9x^2 + 4/9) = -4##

##2x^2 – 4/3x^4 + 8/3x^2 -4/3 = -4##

##-4/3x^4 +14/3x^2 + 8/3 = 0##

Solve using a graphing calculator. If it’s a standard one, like a TI84, use ##y_1 = -4/3x^4 + 14/3x^2 + 8/3 = 0## and ##y_2 = 0##, and press “calc” followed by “intersect”.

This will give you real roots of ##2## and ##-2##. All that is left to do is solve for ##y##.

##y = -2/3(2)^2 + 2/3″ AND “-2/3(-2)^2 + 2/3##

##y = -8/3 + 2/3″ AND “-8/3 + 2/3##

##y = -2″ AND ” -2##

Hence, our solution set is ##{2, -2}## and ##{ -2, -2}##.

Use the following practice exercises to develop your comfort with the skills dealt with in this answer.

Practice exercises:

1. Find the real values of ##x## and ##y## that satisfy the following systems of equations.

a) ##2x – 3y = 4, x + 2y = 9##

b) ##3x + y = -2, x^2 = y##

c) ##2x^2 – 3y^2 = -10, x^2 – 2x + 3y = 5##

Hopefully this helps, and good luck!

I’m not sure where exactly you mean “final” in the solving process. So, I’ve prepared a couple of problems that I will through slowly and carefully, showing all the steps to the final answer.

Example 1: Solve the following system of equations-##2x + y = 5, 3x + 2y = 9##

Since we want to solve with substitution, we must solve for one variable in one of the equations. I think it would be easiest to solve for ##y## in the first equation.

##y = 5 – 2x##

We can now substitute into the other equation:

##3x + 2(5 – 2x) = 9##

##3x + 10 – 4x = 9##

##-x = -1##

##x = 1##

We must now find the value of ##y##. This is found by inserting ##x = 1## into one of the equations and solving for ##y##.

##y = 5 – 2x##

##y = 5 – 2(1)##

##y = 3##

Hence, our solution set is ##{1, 3}##.

Example 2: Find all real values of ##x## and ##y## that satisfy the following system of equations: ##3y = -2x^2 + 2, 2x^2 – 3y^2 = -4##

Once again, as with the last example, we need to solve for one of the variables in one of the equations. It looks easiest to isolate the ##y## in the first equation, however solving this equation won’t be as neat as solving the previous one.

##y = -2/3x^2 + 2/3##

We can now substitute into equation 2.

##2x^2 – 3(-2/3x^2 + 2/3)^2 = -4##

##2x^2 – 3(4/9x^4 – 8/9x^2 + 4/9) = -4##

##2x^2 – 4/3x^4 + 8/3x^2 -4/3 = -4##

##-4/3x^4 +14/3x^2 + 8/3 = 0##

Solve using a graphing calculator. If it’s a standard one, like a TI84, use ##y_1 = -4/3x^4 + 14/3x^2 + 8/3 = 0## and ##y_2 = 0##, and press “calc” followed by “intersect”.

This will give you real roots of ##2## and ##-2##. All that is left to do is solve for ##y##.

##y = -2/3(2)^2 + 2/3″ AND “-2/3(-2)^2 + 2/3##

##y = -8/3 + 2/3″ AND “-8/3 + 2/3##

##y = -2″ AND ” -2##

Hence, our solution set is ##{2, -2}## and ##{ -2, -2}##.

Use the following practice exercises to develop your comfort with the skills dealt with in this answer.

Practice exercises:

1. Find the real values of ##x## and ##y## that satisfy the following systems of equations.

a) ##2x – 3y = 4, x + 2y = 9##

b) ##3x + y = -2, x^2 = y##

c) ##2x^2 – 3y^2 = -10, x^2 – 2x + 3y = 5##

Hopefully this helps, and good luck!