You move one of the like terms to the other side using algebraic operations. For instance...
3x + 5y = 7x + 3
... and you want to solve for x ...
... first, subtract 3x from both sides ...
3x - 3x + 5y = 7x - 3x + 3
5y = 4x + 3
... do I need to explain further???
For the product to be zero, any of the factors must be zero, so you solve, separately, the two equations: sin x = 0 and: cos x = 0 Like many trigonometric equations, this will have an infinity of solutions, since sine and cosine are periodic functions.
no,, to be like term they must have the same variable,, such as 4x and 9x.. or 1y and 300y...
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The system of equations can have zero solutions, one solution, two solutions, any finite number of solutions, or an infinite number of solutions. If it is a system of LINEAR equations, then the only possibilities are zero solutions, one solution, and an infinite number of solutions. With linear equations, think of each equation describing a straight line. The solution to the system of equations will be where these lines intersect (a point). If they do not intersect at all (or maybe two of the lines intersect, and the third one doesn't) then there is no solution. If the equations describe the same line, then there will be infinite solutions (every point on the line satisfies both equations). If the system of equations came from a real world problem (like solving for currents or voltages in different parts of a circuit) then there should be a solution, if the equations were chosen properly.
No. Like terms should contain the same variable or variables, raised to the same powers. Like terms are those that can be combined by addition or subtraction.
First, if you have two equations like, for instance .90x+2000 and .40x+3500 then you would set them equal to each other like this ( .90x+2000=.40x+3500). Then you would solve for x by simplifying the equations as far down as you can.
There isn't a universal way to do this, just like there isn't a universal way to solve nonlinear equations in one variable. A good place to start, however, would be to attempt to solve an equation for one of the variables, in terms of the other two. If you substitute that into the other equations, you will then have a system of two equations in two variables. Do this again, and you'll have a single variable equation that you'll hopefully know how to solve.
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Sorry, that is not an equation, since you have not told us what those terms equal. An equation would look like this: x2 + 6x -18 = 0 (but it doesn't have to equal zero, that is just an example of what it might be). Until we have an actual equation, we cannot solve for x. Also note: you can go to the site wolframalpha.com and they can solve equations and graph them for you. Very convenient.
Start by collecting like terms...
To solve equations like x2+a=0 with a>0
The answer to the equation is p is equal to nine. You solve the equation by putting like terms together and then solving for p.
You substract 6 to -8 so its equal to y over 6 equal -2 and the you multiply -2 by 6! So the y equal -12...
This problem actually has no solution. Any two numbers that will add to 6.5 will be too small to produce a multiple equal to 64. However, for problems like this in general, writing the question as a formula might help you solve it. 1. Assume the numbers are x and y. 2. Write two equations (two unknowns and two equations gives you enough information to solve). -The first part, "two numbers added equal 6.5" can be written as x + y = 6.5 -The second part, "two numbers multiplies equal 64" as xy = 64 3. Manipulate the equations to solve. For example: y = 64/x x + y = x + (64/x) = 6.5 4. Solve for x, and then use x to find y using either of the original formulas.
combine like terms order of operations () 2 X / + - and that's it.
Joint variation equations are equations that have a variable equal to the product of two or more other variables and usually a coefficient. For example, an equation like x=2yz.
You can solve a system of equations using a few methods: elemination and subsitution. In the elemination method you would multiply each of your equations by a factor that would cause one of the variables to be eleminated. ex: If you had 2x-3y=5 and 3x+6y=4 you would multiply the first equation by a factor of 2 (distributing the 2 to each term), giving you a new equation of 4x-6y=10. Now your y terms would cancel. You line up the equations like: 3x+6y=4 4x-6y=10 you simply cancel the y terms and add the others giving you: 7x=14 you then solve for x (in this case divide by 7) x=2 now you plug in x=2 into one of the first equations and solve 3x+6y=4 3(2)+6y=4 plug in x 6+6y=4 multiply 6y=-2 add like terms y=-1/3 solve for y To solve using the subsitution method you would take an equation and solve for one variable then plug that into the other equation. ex. (using the same equations as before) 2x-3y=5 2x=3y+5 x=(3/2)y+(5/2) 3x+6y=4 3((3/2)y+(5/2))+6y=4 plug in for x (9/2)y+(15/2)+6y=4 distribute the 3 (21/2)y=(-7/2) add like terms y=(-1/3) solve for y you would then plug in y and solve for x 2x-3(-1/3)=5 2x+1=5 2x=4 x=2 Hope this helps!