Answer:
a. (7, 27) Polly was incorrect.
b. (3, 1) Polly was correct.
Explanation:
a.
y = 4x - 1
2x - y = -13
We can solve this by substitution. Our first equation gives us y. Plug that into our second equation for each instance of y.
2x - y = -13
2x - (4x - 1) = -13
First distribute the negative sign across the parentheses.
2x - 4x + 1 = -13
Combine like terms.
-2x + 1 = -13
-2x = -14
Divide both sides by -2.
x = 7
Now that we know x, plug its value into one of our original equations.
y = 4x - 1
y = 4(7) - 1
Solve.
y = 28 - 1
y = 27
Our solution is (7, 27).
Let's check our answer by plugging both values into one of our original equations.
2x - y = -13
2(7) - 27 = -13
Solve.
14 - 27 = -13
-13 = -13
Your solution is correct.
Polly's solution was (-2, -9), which is incorrect.
b.
3x + 6y = 15
-2x + 3y = -3
Let's solve this by elimination. To solve by elimination means to eliminate a variable from the equation.
We have 3x & -2x and 6y & 3y. Notice that 6 and 3 can be simplified easily. To eliminate y, we want to multiply each factor of the second equation by -2.
-2x + 3y = -3
-2(-2x + 3y = -3)
4x - 6y = 6
Write the equations together.
3x + 6y = 15
4x - 6y = 6
Now we can see that the y variable will cancel out.
Apply the elimination method.
3x + 6y = 15
4x - 6y = 6
________
7x = 21
Divide both sides by 7.
x = 3
Now that we know x, plug its value into one of our original equations.
3x + 6y = 15
3(3) + 6y = 15
Solve.
9 + 6y = 15
6y = 15 - 9
6y = 6
y = 1
Our solution is (3, 1).
Let's check our answer by plugging both values into one of our original equations.
-2x + 3y = -3
-2(3) + 3(1) = -3
Solve.
-6 + 3 = -3
-3 = -3
Your solution is correct.
Polly's solution was (3, 1), so she was correct.
Hope this helps!