How to find the solution for a system of equations - Algebra

The question is asking the student to solve the linear set of equations ultimately by isolating and .

There are a few ways a student could choose to answer this.

One may see immediately and realize to eliminate the result from the second equation one would subtract as follows.

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Other students may choose to subtract the first equation twice from the second equation then subsequently solve for y.

To solve this system of equations, multiply each equation by an integer that would allow us to cancel one variable. We then substitute that value back into either original equation to find the other variable. In this case we will cancel the x variable by multiplying each equation by 3 and 4 respectively.

Multiply equation 1 by 3 and equation 2 by 4:

Add the new equations:

The result is

Solve for y:

Now substitute this value back into equation 1:

For the second equation, solve for in terms of .

Plug this value of y into the first equation.

Solve the second equation for y:

x - 2y = 4

-2y = 4 - x

y = -2 + x/2

Plug this into the first equation:

3x + 2(-2 + x/2) = 8

Solve for x:

3x - 4 + x = 8

4x = 12

x = 3

Plug this into the second equation to get a value for y:

3 - 2y = 4

2y = -1

y = -0.5

Set the two equations equal to one another:

2x - 2 = 3x + 6

Solve for x:

x = -8

Plug this value of x into either equation to solve for y. We'll use the top equation, but either will work.

y = 2 * (-8) - 2

y = -18

Since the object in question is a cube, each of its sides must be the same length. Therefore, to get a volume of , each side must be equal to the cube root of , which is cm.

We can then set each expression equal to .

The first expression can be solved by either or , but the other two expressions make it evident that the solution is .

We can solve this problem by setting up a simple system of equations. First, we want to change the equations so one variable can cancel out. Multiplying the first equation by 2 and the second equation by 3 gives us a new system of and . These equations add up to or . Plugging in 7 for in either of the original two equations shows us that is equal to 1 and the point is .

The most simple method for solving systems of equations is to transform one of the equations so it allows for the canceling out of a variable. In this case, we can multiply by to get .

Then, we can add to this equation to yield , so .

We can plug that value into either of the original equations; for example, .

So, as well.

rearranges to

and

, so

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

Now, substitute 2 for in the first equation:

The solution is

By solving one equation for , and replacing in the other equation with that expression, you generate an equation of only 1 variable which can be readily solved.

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.

Solve the second equation for , allowing us to solve using the substitution method.

Substitute for in the first equation, and solve for .

Now, substitute for in either equation; we will choose the second. This allows us to solve for .

Now we can write the solution in the notation , or .

To solve this system of equations, use substitution. First, convert the second equation to isolate .

Then, substitute into the first equation for .

Combine terms and solve for .

Now that we know the value of , we can solve for using our previous substitution equation.

To solve this system of equations, we must first eliminate one of the variables. We will begin by eliminating the variables by finding the least common multiple of the variable's coefficients. The least common multiple of 3 and 2 is 6, so we will multiply each equation in the system by the corresponding number, like

.

By using the distributive property, we will end up with

Now, add down each column so that you have

Then you solve for and determine that .

But you're not done yet! To find , you have to plug your answer for back into one of the original equations:

Solve, and you will find that .

When we add the two equations, the and variables cancel leaving us with:

which means there is no solution for this system.

When we add the two equations, the variables cancel leaving us with:

Solving for we get:

We can then substitute our value for into one of the original equations and solve for :

Multiply the bottom equation by 5, then add to the top equation:

Multiply the top equation by :

Now add: