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So, just get started with a step that’s most convenient and simple. ⇨ y = –4 + x Now, substituting y in equation 2, we have:Īs you can see, the solution in both the cases is the same. Isolate y First Isolating y from equation 1, we have: Now, all we need to do is just substitute the value of x into this. We have rearranged the equation 2 and have an expression for y all set! We can put this x-value back into any of the given equations and solve for y. Now substituting y in the first equation, we have: Solving the second equation for y, we get: Let’s choose the second equation this time. Solve 4x – 6y = –16 and 8x + 2y = 24 using the substitution method.
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Thus, the solution of the linear system is (1, 7). Now the last step! We need to plug this value of x back in one of the equations. Now, let’s substitute the expression for y in the second equation.Īpplying the distributive property, we have: In the above system, the variable y is isolated. Let's understand the steps with this linear system. ✯ Plug the value back into one of the equations to solve for the variable initially isolated. ✯ Substitute the expression so obtained into the other equation to solve for the other variable. ✯ Rearrange an equation to isolate one of the variables on one side. The substitution method involves three steps. Solving Systems of Equations by Substitution In this lesson, we are going to talk you through solving a system of 2 linear equations in 2 variables. The method can also be used to find the solution of a system of three or more equations in three or more variables, but it takes longer. Substitution is the quickest method of solving a system of two equations in two variables.