Example Card: Solving a System Using Substitution
Solve systems of equations using substitution in Grade 9 algebra. Substitute one variable expression into the second equation to find the intersection point of two lines.
Key Concepts
Sometimes graphing isn't precise. Let's use algebra to find the exact solutions. The second key idea is solving using substitution, which gives us precise coordinates.
Example Problem Solve the system using substitution: $y = x^2 + 3x 2$ and $y = 3x + 7$.
Step by Step 1. Since both equations are equal to $y$, we can set their right sides equal to each other. This is the substitution step. $$x^2 + 3x 2 = 3x + 7$$ 2. To solve for $x$, we need to get the equation into the standard quadratic form ($ax^2+bx+c=0$). We can do this by adding the expression $ 3x 7$ to both sides. $$x^2 9 = 0$$ 3. We should recognize the left side of the equation, $x^2 9$, as a difference of two squares. $$(x+3)(x 3) = 0$$ 4. Now, we use the Zero Product Property and set each factor to zero to find the two possible values for $x$. $$x+3=0 \text{ and } x 3=0$$ $$x = 3 \text{ and } x = 3$$ 5. We have two $x$ values, so we must find the corresponding $y$ value for each. We can use the simpler linear equation, $y = 3x + 7$. 6. First, substitute $x = 3$: $$y = 3( 3) + 7 \Rightarrow y = 9 + 7 \Rightarrow y = 2$$ 7. Next, substitute $x = 3$: $$y = 3(3) + 7 \Rightarrow y = 9 + 7 \Rightarrow y = 16$$ 8. The solutions are the ordered pairs $( 3, 2)$ and $(3, 16)$.
Common Questions
How does the substitution method work for systems of equations?
Solve one equation for one variable, substitute that expression into the other equation to get a single-variable equation, solve it, then back-substitute to find the second variable.
When should you use substitution vs. elimination?
Use substitution when one equation already has a variable isolated or has a coefficient of 1. Elimination works better when coefficients are larger and can be easily matched.
How do you verify a solution to a system of equations?
Substitute both values back into both original equations. If both equations are true with those values, the solution is correct.