What is the final step of completing a solve by substitution problem?

Question

What is the final step of completing a solve by substitution problem?

1 Answer

Impact of this question

4946 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-20T16:48:18

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

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

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 $y$ $y$ in the first equation.

$y = 5 - 2 x$ $y = 5 - 2x$

We can now substitute into the other equation:

$3 x + 2 (5 - 2 x) = 9$ $3x + 2(5 - 2x) = 9$

$3 x + 10 - 4 x = 9$ $3x + 10 - 4x = 9$

$- x = - 1$ $-x = -1$

$x = 1$ $x = 1$

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

$y = 5 - 2 x$ $y = 5 - 2x$

$y = 5 - 2 (1)$ $y = 5 - 2(1)$

$y = 3$ $y = 3$

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

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

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 $y$ $y$ in the first equation, however solving this equation won't be as neat as solving the previous one.

$y = - \frac{2}{3} x^{2} + \frac{2}{3}$ $y = -2/3x^2 + 2/3$

We can now substitute into equation #2.

$2 x^{2} - 3 {(- \frac{2}{3} x^{2} + \frac{2}{3})}^{2} = - 4$ $2x^2 - 3(-2/3x^2 + 2/3)^2 = -4$

$2 x^{2} - 3 (\frac{4}{9} x^{4} - \frac{8}{9} x^{2} + \frac{4}{9}) = - 4$ $2x^2 - 3(4/9x^4 - 8/9x^2 + 4/9) = -4$

$2 x^{2} - \frac{4}{3} x^{4} + \frac{8}{3} x^{2} - \frac{4}{3} = - 4$ $2x^2 - 4/3x^4 + 8/3x^2 -4/3 = -4$

$- \frac{4}{3} x^{4} + \frac{14}{3} x^{2} + \frac{8}{3} = 0$ $-4/3x^4 +14/3x^2 + 8/3 = 0$

Solve using a graphing calculator. If it's a standard one, like a TI84, use $y_{1} = - \frac{4}{3} x^{4} + \frac{14}{3} x^{2} + \frac{8}{3} = 0$ $y_1 = -4/3x^4 + 14/3x^2 + 8/3 = 0$ and $y_{2} = 0$ $y_2 = 0$ , and press "calc" followed by "intersect".

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

$y = - \frac{2}{3} {(2)}^{2} + \frac{2}{3} AND - \frac{2}{3} {(- 2)}^{2} + \frac{2}{3}$ $y = -2/3(2)^2 + 2/3" AND "-2/3(-2)^2 + 2/3$

$y = - \frac{8}{3} + \frac{2}{3} AND - \frac{8}{3} + \frac{2}{3}$ $y = -8/3 + 2/3" AND "-8/3 + 2/3$

$y = - 2 AND - 2$ $y = -2" AND " -2$

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

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

Practice exercises:

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

a) $2 x - 3 y = 4, x + 2 y = 9$ $2x - 3y = 4, x + 2y = 9$

b) $3 x + y = - 2, x^{2} = y$ $3x + y = -2, x^2 = y$

c) $2 x^{2} - 3 y^{2} = - 10, x^{2} - 2 x + 3 y = 5$ $2x^2 - 3y^2 = -10, x^2 - 2x + 3y = 5$

Hopefully this helps, and good luck!

Science

Math

Humanities

... and beyond

What is the final step of completing a solve by substitution problem?

1 Answer

Related questions

Impact of this question