How can I solve the system of equations y = x^2 + 3x – 7 and 3x – y = 2 using substitution?

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To solve the system of equations using substitution, we start with the two equations:

1. y = x^2 + 3x - 7
2. 3x - y = 2

We can substitute the expression for y from the first equation into the second equation.

1. First, rewrite the second equation:

3x - y = 2

2. Now, substitute y with x^2 + 3x – 7: 

3x - (x^2 + 3x - 7) = 2

3. Simplify the equation:

3x - x^2 - 3x + 7 = 2
-x^2 + 7 = 2

4. Next, isolate x:

-x^2 + 7 - 2 = 0
-x^2 + 5 = 0
x^2 = 5

5. Now, solve for x:

x = ±√5

6. Substitute the values of x back into the first equation to find the corresponding y values:

y = (√5)^2 + 3(√5) - 7
y = 5 + 3√5 - 7
y = 3√5 - 2
y = (−√5)^2 + 3(−√5) - 7
y = 5 - 3√5 - 7
y = -3√5 - 2

7. Therefore, the solutions for the system of equations are:

  • (√5, 3√5 – 2)
  • (−√5, -3√5 – 2)

In summary, we found the values of x and y using substitution, providing us with two solution pairs for the equations.

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