Solving Equations - Algebra 2

Write the second equation in slope-intercept form

This is the same as the first equation.

The two equations represent the same line, therefore there are an infinite number of solutions.

Distribute the x through the parentheses:

x2 –2x = x2 – 8

Subtract x2 from both sides:

–2x = –8

Divide both sides by –2:

x = 4

Given the equation,

and

Plug in for to the equation,

Solve and simplify.

Multiply both sides by 3:

Distribute:

Subtract from both sides:

Add the terms together, and subtract from both sides:

Divide both sides by :

Simplify:

Plug in the value for .

Simplify

Subtract

On the equation, replace x with 2 and then simplify.

Equation 1:

Equation 2:

Equation 3:

Adding the terms of the first and second equations together will yield .

Then, add that to the third equation so that the y and z terms are eliminated. You will get .

This tells us that x = 1. Plug this x = 1 back into the systems of equations.

Now, we can do the rest of the problem by using the substitution method. We'll take the third equation and use it to solve for y.

Plug this y-equation into the first equation (or second equation; it doesn't matter) to solve for z.

We can use this z value to find y

So the solution set is x = 1, y = 2, and z = –5/3.

rearranges to

and

, so

To solve this problem we can first add to each side of the equation yielding

Then we take the square root of both sides to get

Then we calculate the square root of which is .

In the second equation, you can substitute for from the first.

Now, substitute 2 for in the first equation:

The solution is

Isolate in the first equation.

Plug into the second equation to solve for .

Plug into the first equation to solve for .

Now we have both the and values and can express them as a point: .

Add the equations together.

Put the terms together to see that .

Substitute this value into one of the original equaitons and solve for .

Now we know that , thus we can find the sum of and .

To find the point where these two lines intersect, set the equations equal to each other, such that is substituted with the side of the second equation. Solving this new equation for will give the -coordinate of the point of intersection.

Subtract from both sides.

Divide both sides by 2.

Now substitute into either equation to find the -coordinate of the point of intersection.

With both coordinates, we know the point of intersection is . One can plug in for and for in both equations to verify that this is correct.

To solve for , you must isolate it from the other variables. Start by adding to both sides to give you . Now, you need only to divide from both sides to completely isolate . This gives you a solution of .

First, you must multiply the left side of the equation using the distributive property.

This gives you .

Next, subtract from both sides to get .

Then, divide both sides by to get .

Combine like terms on the left side of the equation:

Use the distributive property to simplify the right side of the equation:

Next, move the 's to one side and the integers to the other side:

To solve for , you must isolate it so that all of the other variables are on the other side of the equation. To do this, first subtract from both sides to get . Then, divide both sides by to get .

This is a quadratic equation. We can solve for either by factoring or using the quadratic formula. Since this equation is factorable, I will present the factoring method here.

The factored form of our equation should be in the format .

To yield the first value in our original equation (), and .

To yield the final term in our original equation (), we can set and .

Now that the equation has been factored, we can evaluate . We set each factored term equal to zero and solve.

Multiply the top equation by 3 on both sides, then add the second equation to eliminate the terms: