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The basic method is the same as for other types of equations: you need to isolate the variable ("x", or whatever variable you need to solve for). In the case of radical equations, it often helps to square both sides of the equation, to get rid of the radical. You may need to rearrange the equation before squaring. It is important to note that when you do this (square both sides), the new equation may have solutions which are NOT part of the original equation. Such solutions are known as "extraneous" solutions. Here is a simple example (without radicals): x = 5 (has one solution, namely, 5) Squaring both sides: x squared = 25 (has two solutions, namely 5, and -5). To protect against this situation, make sure you check each "solution" of the modified equation against the original equation, and reject the solutions that don't satisfy it.
You first need to isolate one variable in one of the equations and then substitute that value into the other equation and solve for the remaining variable. Take the value you just got and plug it in to the other equation for the appropriate variable. Solve for the first variable that you isolated. Example: 2x-y=3, 4x+2y=6 2x-y=3 Isolate y. +y +y 2x=y+3 -3 -3 y=2x-3 If y=2x-3 then substitute 2x-3 in for y in the other equation. 4x+2y=6 4x+2(2x-3)=6 Distribute 4x+4x-6=6 Simplify 8x-6=6 +6 +6 8x=12 Divide both sides by 8 to isolate x. x=12/8 Simplify x=3/2 Now substitute 3/2 in for x in the first equation. 2x-y=3 2(3/2)-y=3 Again, distribute. 6/2-y=3 Simplify 3-y=3 Isolate y. -3 -3 y=0 (It can't be -y=0, because you can't have -0) So, x=3/2 and y=0!
By knowing how to use the quadratic equation formula.
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Start with a quadratic equation in the form � � 2 � � � = 0 ax 2 +bx+c=0, where � a, � b, and � c are constants, and � a is not equal to zero ( � ≠ 0 a =0).
The basic method is the same as for other types of equations: you need to isolate the variable ("x", or whatever variable you need to solve for). In the case of radical equations, it often helps to square both sides of the equation, to get rid of the radical. You may need to rearrange the equation before squaring. It is important to note that when you do this (square both sides), the new equation may have solutions which are NOT part of the original equation. Such solutions are known as "extraneous" solutions. Here is a simple example (without radicals): x = 5 (has one solution, namely, 5) Squaring both sides: x squared = 25 (has two solutions, namely 5, and -5). To protect against this situation, make sure you check each "solution" of the modified equation against the original equation, and reject the solutions that don't satisfy it.
You first need to isolate one variable in one of the equations and then substitute that value into the other equation and solve for the remaining variable. Take the value you just got and plug it in to the other equation for the appropriate variable. Solve for the first variable that you isolated. Example: 2x-y=3, 4x+2y=6 2x-y=3 Isolate y. +y +y 2x=y+3 -3 -3 y=2x-3 If y=2x-3 then substitute 2x-3 in for y in the other equation. 4x+2y=6 4x+2(2x-3)=6 Distribute 4x+4x-6=6 Simplify 8x-6=6 +6 +6 8x=12 Divide both sides by 8 to isolate x. x=12/8 Simplify x=3/2 Now substitute 3/2 in for x in the first equation. 2x-y=3 2(3/2)-y=3 Again, distribute. 6/2-y=3 Simplify 3-y=3 Isolate y. -3 -3 y=0 (It can't be -y=0, because you can't have -0) So, x=3/2 and y=0!
To determine the value of a radical, you can simplify it by factoring the expression under the radical into perfect squares or other known values. Another approach is to estimate the value by identifying perfect squares close to the radical's value, allowing you to approximate. Additionally, you can use a calculator for precise values or apply numerical methods such as the Newton-Raphson method for more complex radicals. Lastly, graphing the function can provide a visual representation of the radical’s value.
You can not increase a "Method". A method is a way of doing something. You could change your method, improve your method, simplify tour method, but NOT "increase" it.
An expression cannot be solved. It may be simplified, but that is not the same thing. And how it is simplified depends on its form. An equation, or inequality, can be solved but, again, the method depends on its nature.
To solve this system of equations, we can use the method of substitution or elimination. Let's use the substitution method. From the second equation, we can express y as y = 55 - 4x. Substitute this expression for y in the first equation: 7x - 5(55 - 4x) = 76. Simplify this equation to solve for x. Then, substitute the value of x back into one of the original equations to find the value of y.
To solve the equation (2x + 3y = 12) and (5x - y = 13), you can use the substitution or elimination method. First, express one variable in terms of the other from one equation, and substitute it into the second equation. For example, from the first equation, you can isolate (y) as (y = \frac{12 - 2x}{3}) and substitute this into the second equation to find (x). After finding (x), substitute it back to find (y).
You can combine equivalent terms. You should strive to put the equation in the form ax2 + bx + c = 0. Once it is in this standard form, you can apply the quadratic formula, or some other method, to solve it.
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His method to figure out the difficult algebra equation was sucessful.