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

Use elimination to solve the solution:

Explanation

For elimination you need to get one variable by itself by cancelling the other out. In this equation this is best done by getting rid of . You can multiply whichever equation you would like to, but multiply it by to get

then add the equations together

which, simplified, is

divied by to get

Then plug back into any equation for the x value

Solve for to get

Solve this system of equations:

Explanation

We can rewrite the first equation as:

If we substitute this new value for into the second equation we get:

Simplify.

Combine like terms

Solve for

Now substitute this value into either of the original equations:

In the standard coordinate plane, slope-intercept form is defined for a straight line as , where is the slope and is the point on the line where .

Give the coordinates at which the following lines intersect:

$(\frac{14}{3}, 11)$

$(8, \frac{5}{3})$

$(11, \frac{14}{3})$

$(\frac{5}{3}, 8)$

The two lines do not intersect.

Explanation

The first step is to set the two equations equal to each other, since at the point of intersection, they will be. To do this more easily, convert each equation into slope-intercept form.

First equation:

State equation

Add to both sides.

$y = \frac{3}{2}x +4$ Divide both sides by .

Second equation:

State equation

Symmetric Property of Identity

Now, since each equation equals , the equations also equal each other (for the point of intersection). By solving for , therefore, we can

$\frac{3}{2}x + 4 = 3x-3$ State equation.

$\frac{3}{2}x + 7 = 3x$ Add to both sides.

$7 = \frac{3}{2}x$ Subtract $\frac{3}{2}x$ from both sides.

$\frac{14}{3} = x$ Divide both sides by $\frac{3}{2}$ (or multiply both sides by $\frac{2}{3}$ ).

So, the -coordinate of our intersection is $x= \frac{14}{3}$ . To find the -coordinate, plug this result back into one of the original equations.

State your chosen equation.

$y= 3(\frac{14}{3})-3$ Substitute the value of .

Multiply.

Subtract.

So, the coordinates where the two lines intersect are $(\frac{14}{3}, 11)$ .

Without drawing a graph of either equation, find the point where the two lines intersect.

Line 1 :

Line 2 :

Explanation

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.

$\frac{2x}{2} = \frac{-2}{2}$

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.

In the standard coordinate plane, slope-intercept form is defined for a straight line as , where is the slope and is the point on the line where .

Give the coordinates at which the following lines intersect:

Explanation

The first step is to set the two equations equal to each other, since at the point of intersection, they will be. To do this more easily, convert each equation into slope-intercept form.

First equation:

State equation

Add to both sides.

Second equation:

State equation

Divide both sides by .

Subtract from both sides.

Now, since each equation equals , the equations also equal each other (for the point of intersection). By solving for , therefore, we can

State equation.

Subtract from both sides.

Divide both sides by .

So, the -coordinate of our intersection is . To find the -coordinate, plug this result back into one of the original equations.

State your chosen equation.

Substitute the value of .

So, the coordinates where the two lines intersect are .

Find the solution to this system of equations:

None of these

Explanation

To find the solution we will multiply by some number to cancel out one variable and then solve for the variable that is left over. This is called the "elimination" method. You may choose to eliminate either variable, but it is prudent to choose the variable that is easier to cancel.

Multiply equation 1 by -2:

$-2(-x+3y=-6)\rightarrow 2x-6y=12$

Add new equation 1 to equation 2:

The result is

Solve for y:

$\mathbf{y=0}$

Now solve for x by plugging in the value for y (we can choose either original equation):

$-2x+0=-12\rightarrow \mathbf{x=6}$

Solve for in the system of equations:

The system has no solution

Explanation

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

Now, substitute 2 for in the first equation:

The solution is

Solve the following system of equation

$(x,y,z)=\left (-\frac{5}{4},-2,-2 \right)$

$(x,y,z)=\left (-2,2,-\frac{5}{4} \right )$

$(x,y,z)=\left (-5,-3,-4} \right )$

$(x,y,z)=\left (-\frac{7}{2},-9,-8 \right )$

Cannot be solved

Explanation

Start with the equation with the fewest variables, $\ 6z=-12$ .

Solve for by dividing both sides of the equaion by 6:

$\frac{6z}{6}=\frac{-12}{6}$

Plug this value into the second equation to solve for :

Subtract 10 from both sides:

Divide by 9:

$\frac{9y}{9}=\frac{-18}{9}$

Plug these and values into the first equation to find :

Combine like terms:

Subtract 2:

Divide by -4:

$\frac{-4x}{-4}=\frac{5}{-4}$

$x=-\frac{5}{4}$

Therefore the final solution is $(x,y,z)=\left (-\frac{5}{4},-2,-2 \right)$ .

What is the sum of and for the following system of equations?

Explanation

Add the equations together.

Put the terms together to see that $x+0=5\ \textup{or}\ x=5$ .

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

Now we know that $x=5\ \textup{and}\ y=2$ , thus we can find the sum of and .

In the standard coordinate plane, slope-intercept form is defined for a straight line as , where is the slope and is the point on the line where .

Give the coordinates at which the following lines intersect:

$y = \frac{3}{2}x + 7$

The two lines do not intersect.

$(\frac{7}{2}, -\frac{3}{2})$

$(-\frac{3}{2}, \frac{7}{2})$

$(3, \frac{7}{2})$

$(\frac{7}{2}, 3)$

Explanation

The first step is to set the two equations equal to each other, since at the point of intersection, they will be. To do this more easily, convert each equation into slope-intercept form.

First equation:

$y = \frac{3}{2}x + 7$ Already in slope-intercept form

Second equation:

State equation

Add to both sides.

$y = \frac{9}{6}x + \frac{14}{6}$ Divide both sides by .

$y = \frac{3}{2}x + \frac{7}{3}$

At this point, note that both equations have idential slopes: $m = \frac{3}{2}$ for both equations, but different -intercepts. Thus, the lines are parallel, and will never touch. We can stop here, but let's prove our theory with algebra by setting the equations equal to one another: