Solve the following system using substitution. Write the solution as an ordered pair, (a, b).
b = a – 4
2a – 3b = -2
okay just equation 1 has b= a - 4 we can plug in a- 4 for b in equation 2 to give you;
2a - 3(a - 4) = -2
2a - 3a +4= -2
-a + 4 = -2
-a = -6
divide by negative 1 to make a postive
a= 6 now you can plug in 6 into either equation to solve for b
but i would use equation 1 since its so short b = 6- 4 , b= 2
(6, 2)
i hope that helps if your confuse about anything with the above solution just ask and ill try to clarify
The only problem is, a=6 b=2 does not satisfy the original equations.
If you work it more carefully, you will wind up with
a=14
b=10
hint: -3(a-4) is not -3a+4
To solve the system of equations using the substitution method, we will start by solving one equation for one variable and then substitute that expression into the other equation.
Let's solve the first equation, b = a - 4, for b:
b = a - 4
Next, we substitute this expression for b in the second equation, 2a - 3b = -2:
2a - 3(a - 4) = -2
Simplifying the equation:
2a - 3a + 12 = -2
-a + 12 = -2
Now, let's isolate the variable:
-a = -2 - 12
-a = -14
Dividing both sides by -1:
a = 14
Now that we have found the value of a, we can substitute it back into the first equation to find b:
b = a - 4
b = 14 - 4
b = 10
Therefore, the solution to the system of equations is (a, b) = (14, 10).