Normally you would need the remainder of the equation to solve it. However, anything times one equals itself, so the answer (in part) is "2y". Remember, you can't solve the equation fully without finding the value of 'y', and you cannot find the value of 'y' unless you know what resides on the other side of the equal sign.
You solve this as follows. You call your number "x", and write:x = .323232... (equation 1) Then you multiply this equation by 100 (a one, followed by two zeros - this is because the length of the period is 2), and get: 100x = 32.323232... (equation 2) Now you can subtract equation 1 from equation 2, and solve for "x". This will give you "x" as a fraction.
Mathematical substitution is the process of using one equation to solve for multiple variables. For example: Equation 1: x + y = 4 Equation 2: 2x + y = 16 Using equation 1, solve for y: y = 4 - x <-- Plug this into equation 2. This is substitution because you are replacing y in equation 2 with what y is equal to in equation 1. 2x + y = 16 ----> 2x + (4 - x) = 16 Now you can solve for x: x + 4 = 16; x = 12 You can then substitute the value of x back into the equation that is solved for y: y = 4 - 12; y = -8 Check both equations: Equation 1: -8 + 12 = 4; 4 = 4 (Correct) Equation 2: 2(12) + (-8) = 16; 24 - 8 = 16; 16 = 16 (Correct) We have successfully used substitution to solve for two different variables, x and y.
Set the equation equal to zero. 3x2 - x = -1 3x2 - x + 1 = 0 The equation is quadratic, but can not be factored. Use the quadratic equation.
Assuming you want the 666... to repeat endlessly, you need to write:x = 7.66666... (equation 1) 10x = 76.66666... equation 2) Then subtract equation 1 from equation 2, and solve for "x". This will give you "x" (your number) as a fraction.
This is a system of two linear equations: x+2y=11 3x-4y=-17 One way to solve this system is to solve for x (or y) in the first equation and then plug it into the second equation to solve for the other variable, which then allows us to solve for our original variable. (1) Take the first equation and solve for x: x=11-2y (2) Plug this value into the second equation: 3x-4y=3(11-2y)-4y=33-10y=-17. We see that y=5. (3) We now plug this into either equation to solve for x: From (1): x=11-2y=11-2(5)=1 We get x=1 and y=5. It's also always a good idea to plug the numbers we found into our original two equations to verify they work. Indeed, x+2y=1+2(5)=1+10=11, and 3x-4y=3(1)-4(5)=3-20=-17.
The equation to solve is given by. |-2 x + 2| -3 = -3 Add 3 to both sides of the equation and simplify. |-2 x + 2| = 0 |-2 x + 2| is equal to 0 if -2 x + 2 = 0. Solve for x to obtain. So, x = 1
You solve this as follows. You call your number "x", and write:x = .323232... (equation 1) Then you multiply this equation by 100 (a one, followed by two zeros - this is because the length of the period is 2), and get: 100x = 32.323232... (equation 2) Now you can subtract equation 1 from equation 2, and solve for "x". This will give you "x" as a fraction.
The solution to the equation 2x plus 2 is 2(x + 1).
2
Call this fraction "x". In that case:x = 0.123123... (equation 1) 1000x = 123.123123... (equation 2) Subtract equation 1 from equation 2, and solve for "x". That gives you the number as a fraction.
Mathematical substitution is the process of using one equation to solve for multiple variables. For example: Equation 1: x + y = 4 Equation 2: 2x + y = 16 Using equation 1, solve for y: y = 4 - x <-- Plug this into equation 2. This is substitution because you are replacing y in equation 2 with what y is equal to in equation 1. 2x + y = 16 ----> 2x + (4 - x) = 16 Now you can solve for x: x + 4 = 16; x = 12 You can then substitute the value of x back into the equation that is solved for y: y = 4 - 12; y = -8 Check both equations: Equation 1: -8 + 12 = 4; 4 = 4 (Correct) Equation 2: 2(12) + (-8) = 16; 24 - 8 = 16; 16 = 16 (Correct) We have successfully used substitution to solve for two different variables, x and y.
To solve the equation X^3 + X^2 = 12, you'll need to find the values of X that satisfy the equation. Here's how you can solve it: Rewrite the equation in standard form, setting it equal to zero: X^3 + X^2 - 12 = 0 This equation is a cubic equation. To solve it, you can try factoring it or use numerical methods such as the Newton-Raphson method. In this case, factoring is a suitable method. First, look for common factors that can be factored out from all the terms. In this case, you can factor out X^2: X^2(X + 1) - 12 = 0 Now, you have a quadratic equation in the form (X^2 - 12)(X + 1) = 0. This equation can be factored further: (X - √12)(X + √12)(X + 1) = 0 You now have three factors: a) X - √12 = 0 b) X + √12 = 0 c) X + 1 = 0 Solve each of these equations for X: a) X - √12 = 0 X = √12 b) X + √12 = 0 X = -√12 c) X + 1 = 0 X = -1 So, the solutions for X in the equation X^3 + X^2 = 12 are X = √12, X = -√12, and X = -1.
in order to solve an equation for x, you must make sure that x is alone, in this case it is not, there is a (-) before it, remember, you have to do the same operation on each side of the equation to make it equivalent to the first equation. if -x=1 x=(-)1 x=-1
Without an equality sign the terms given can't be considered to be an equation.
x2+3x+2=0 (x+2)(x+1)=0 x=-2 or -1
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The equation 6 x-1 18x can be simplified to the answer 112x.