Question

Solve the following system of equations using the elimination method. 2x−3y=14 ,  4x+3y=10

Answer 1

Use the elimination method to solve the linear-quadratic system  y=x^2+2x−4 ,  y=−x−6 .

Answer 2

To solve the system of equations using the elimination method, we can add the two equations together. This will help us eliminate the y term.

Adding the two equations:
(2x - 3y) + (4x + 3y) = 14 + 10
6x = 24

To isolate x, we divide both sides of the equation by 6:
6x/6 = 24/6
x = 4

Now that we have the value of x, we can substitute it back into one of the original equations to solve for y. Let's use the first equation:

2x - 3y = 14
2(4) - 3y = 14
8 - 3y = 14

Subtract 8 from both sides of the equation:
-3y = 14 - 8
-3y = 6

Divide both sides of the equation by -3:
-3y / -3 = 6 / -3
y = -2

So the solution to the system of equations is x = 4 and y = -2.

Answer 3

To solve the system of equations using the elimination method, we'll equate the expressions of y in both equations and solve for x.

Since both equations are already in the form of "y = ...", we can equate the right sides:

x^2 + 2x - 4 = -x - 6

Next, we'll rearrange the equation to bring all terms to one side:

x^2 + 2x + x - 4 + 6 = 0

We combine like terms:

x^2 + 3x + 2 = 0

Now we have a quadratic equation. To solve it, we'll factor or use the quadratic formula. In this case, the equation factors as:

(x + 1)(x + 2) = 0

This gives us two possible solutions:

x + 1 = 0 or x + 2 = 0

Solving these equations separately, we find:

For x + 1 = 0, subtracting 1 from both sides gives x = -1.

For x + 2 = 0, subtracting 2 from both sides gives x = -2.

So the solutions for x are x = -1 and x = -2.

To find the corresponding y-values, we'll substitute each value of x into one of the original equations. Let's use the second equation:

For x = -1:

y = -x - 6
y = -(-1) - 6
y = 1 - 6
y = -5

For x = -2:

y = -x - 6
y = -(-2) - 6
y = 2 - 6
y = -4

So the solutions for the system of equations are x = -1, y = -5 and x = -2, y = -4.