If you mean: y+x = 3 then the solutions are (0, 3) and (3, 0)
The equation that has the solution x = -3, is, precisely:x = -3 If you want anything more fancy, you can add some number (the same number to both sides), multiply by some number (the same number to both sides), etc.
The roots of the equation are -4 and 3. Simplest solution is (x +4 )(x - 3) which multiplies out as x^2 + x - 12 = 0
x = 00 reduces to x = 0/0. Both are undefined, and thus cannot be considered "equal". Other than that, nothing.
For example, if you have (0, 6) or (3, 1). Which of them is a solution to y - 2x = 6? Check (0, 6): y - 2x = 6, substitute 0 for x, and 6 for y into the equation 6 - 2(0) =? 6 6 - 0 =? 6 6 = 6 True, then (0, 6) is a solution. Check (3, 1): y - 2x = 6, substitute 3 for x, and 1 for y into the equation 1 - 2(3) =? 6 1 - 6 =? 6 -5 = 6 False, then (3, 1) is not a solution.
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The equation that has a solution of ( x = -3 ) can be written as ( x + 3 = 0 ). When you solve this equation, you subtract 3 from both sides, leading to ( x = -3 ). Alternatively, any equation that can be manipulated to reach this solution, such as ( 2x + 6 = 0 ), will also have ( x = -3 ) as a solution.
To determine if ( x^3 ) is a solution of the equation ( 3x - 54x = 0 ), we first simplify the equation. The left side simplifies to ( -51x = 0 ), which implies ( x = 0 ) is the only solution. Since ( x^3 ) is not equal to ( 0 ) for any ( x ) other than ( 0 ), ( x^3 ) is not a solution to the equation.
The equation is (x - 1)*(x - 3) = 0 or x2- 4x + 3 = 0The equation is (x - 1)*(x - 3) = 0 or x2- 4x + 3 = 0The equation is (x - 1)*(x - 3) = 0 or x2- 4x + 3 = 0The equation is (x - 1)*(x - 3) = 0 or x2- 4x + 3 = 0
The equation ( x - 3 = 0 ) can be solved by isolating ( x ). By adding 3 to both sides of the equation, we find that ( x = 3 ). Thus, the solution to the equation is ( x = 3 ).
A linear equation with the solution ( x = 3 ) can be expressed as ( x - 3 = 0 ) or simplified to ( x = 3 ). Other examples of linear equations using this solution include ( 2x - 6 = 0 ) and ( 5x + 15 = 30 ). Each of these equations will yield ( x = 3 ) when solved.
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The equation that has the solution x = -3, is, precisely:x = -3 If you want anything more fancy, you can add some number (the same number to both sides), multiply by some number (the same number to both sides), etc.
There are 2 (main) ways you can find the solution to a quadratic equation:Factorisation: If you take, for example, the equation (x-3)(x-2) = 0. Either (x-3) or (x-2) must equal 0 for the equation to be true: you then have x-3 = 0 (i.e. x = 3) and x-2 = 0 (i.e. x = 2). These are your solutions to the equation. If you have an equation in the form (say) x2+x-12 = 0, you can find that this is equivalent to (x-3)(x+4) = 0, so x = 3 or x = -4.The magic formula: If the equation won't factorise, or you can't see how it would, there is a formula you can use to give you both values of x: for any equation ax2+bx+c = 0, x = (-b±√(b2-4ac))/2a. The two values of x come from the ±: one uses a plus and the other a minus.
The roots of the equation are -4 and 3. Simplest solution is (x +4 )(x - 3) which multiplies out as x^2 + x - 12 = 0
X2 - X - 6 = 0what two factors of - 6 add up to - 1 ?(X + 2)(X - 3)============(- 2, 0 ) and (3, 0 )------------------------------solution set of points
To determine if (0, 2) is a solution to the equation ( y = x^2 ), we substitute ( x = 0 ) into the equation. This gives us ( y = 0^2 = 0 ). Since the corresponding ( y ) value is 0 and not 2, (0, 2) is not a solution to the equation ( y = x^2 ).
An equation that has one solution is a linear equation of the form ( ax + b = c ), where ( a \neq 0 ). For example, the equation ( 2x + 3 = 7 ) has one solution: ( x = 2 ). This is because it can be rearranged to isolate ( x ) as a single value.