In the context of solving a system of equations by substitution, a "useless result" like 12 equals 12 indicates that the two equations are actually dependent, meaning they represent the same line or have infinitely many solutions. Instead of finding a unique solution, you end up with a tautology that confirms the equations are equivalent. This suggests that any solution that satisfies one equation will also satisfy the other, leading to an infinite set of solutions rather than a single point of intersection.
a=3 is the answer to 12a=48
Solving equations in two unknowns requires two independent equations. Since you have only one equation there is no solution.
Equations: x -y = 2 and x^2 -4y^2 = 5 By combining the equations into a single quadratic equation in terms of y and solving it: y = 1/3 or y = 1 By means of substitution the points of intersection are at: (7/3, 1/3) and (3, 1)
Solving equations in two unknowns requires two independent equations. Since you have only one equation there is no solution.
you need 2 equations to solve for 2 variables
6
X = 135 and y = 15 Solved by addition and substitution
a=3 is the answer to 12a=48
Solving equations in two unknowns requires two independent equations. Since you have only one equation there is no solution.
(2,3)
Equations: x -y = 2 and x^2 -4y^2 = 5 By combining the equations into a single quadratic equation in terms of y and solving it: y = 1/3 or y = 1 By means of substitution the points of intersection are at: (7/3, 1/3) and (3, 1)
Solving equations in two unknowns requires two independent equations. Since you have only one equation there is no solution.
Solving equations in two unknowns requires two independent equations. Since you have only one equation there is no solution.
Solving equations in two unknowns requires two independent equations. Since you have only one equation there is no solution.
you need 2 equations to solve for 2 variables
-4m = 24 m = -6
three things: 1) that the value of 4 is equal to the value of 4. 2) you did not obtain any revealing information. 3) your strategy for solving that system of equations was not good.