1. Algebra Practice Problem
The Solution

2. The problem!
Find the numbers for X and Y which satisfies the following simultaneous equations:-
-5x + y = -2 (i)
5x + 4y = 17 (ii)

3. Method 1 - Substitution
The strategy here is to make the problem easier.
If there was only ONE variable in the equation then the problem is easier. But remember you need to know how to solve the easier problem

4. Solution -5x + y = -2 5x + 4y =17 Make x the subject of eqn (i)
-5x +y -y = -2-y
eqn (i)
eqn (ii)
“Undo” adding the y to -5x. Carry out same operation on RHS of equation

5. -5x + 0 = -2 -y
-5x/-5 = (-2-y)/-5
1x = (2+y)/5
“Undo” the multiplying of x by the -5 using the INVERSE operation of multiply i.e. divide. Do this to both sides of the equation

6. So now that we have a value for x we can substitute it into eqn(ii)
x = (2+y)/5
5((2+y)/5)+4y =17
This is an equation in one variable. We have REDUCED the problem

7. Solving the single variable equation! :-}
5((2+y)/5) +4y =17
2 + y + 4y = 17
2 + 5y = 17
5y =
5y + 0 = 15
5y/5 = 15/5
1y = 3
y = 3
5 divide 5 = 1
Gather like TERMS
Make y the subject of the equation
“Undo” the operations that are carried out on y i.e. carry out INVERSE operations.

8. Back substitute y into eqn(i)
-5x + (3) = -2
-5x = -2 -3
-5x + 0 = -5
-5x/-5 = -5/-5
1x = 1
x =1

9. State the solution
x =1; y =3 or
(x,y) = (1,3)

10. Check the solution
Substitute the values (1,3) for (x,y) into eqns (i) and (ii)
-5(1)+(3) = -2 eqn (i)
5(1) + 4(3) = 17 eqn (ii)
EVALUATE the equations
= -2 true!
= 17 true!