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Are there any variables already isolated?
(4,10), see solution.
Since the y in the first equation is already isolated, the easiest way to solve the system is to use the Substitution Method. Let's substitute y=2.5x into the second equation.
(II): y= 2.5x
(II): Multiply
(II): Add terms
(II): .LHS /8.=.RHS /8.
Now that we have found a solution for x, we can substitute that into our first equation to solve for y.
The solution to the system — the point of intersection — is (4,10).
We can see that lines intersect at point (2,4), this is the solution to the system of equations. Next, we will focus on Substitution Method. Let's look at another example system of equations. y=2x y+x=6 To solve this system we substitute 2x for y in the equation (II) and solve it for x.
(II): y= 2x
(II): Add terms
(II): .LHS /3.=.RHS /3.
We found the solution for x. Now, we need to substitute that solution into the equation (I) and solve for y.
(I): x= 2
(I): Multiply
The solution to this system of equations is (2,4). Finally, we will consider the Elimination Method. Let's take a look at an example system of equations. 2y+3x=4 & (I) 4y-3x=6 & (II) To solve this system using the Elimination Method, we will add equation (I) to equation (II) to eliminate one of the variables.
(II): Add (I)
(II): Add terms
(II): .LHS /6.=.RHS /6.
We found the solution for y. Now, we will substitute 2 for y in the first equation and solve it. Let's do it!
(I): y= 2
(I): Multiply
(I): LHS-4=RHS-4
(I): .LHS /3.=.RHS /3.