In the , we saw, how to form and solve linear equations containing just one variable. Today, we will see how to solve equations containing two variables.
Let us begin by answering the following question:
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In the given problem, of the given options,
$3$
values satisfy the equation. But, totally how many values, do you think are there, that will satisfy the equation.
Actually there are
$infinite$
possible values of $x$ and $y$ that will satisfy the given equation. In other words, with the given information there is no way we can solve the given equation to find a single solution, more commonly known as a
$unique\ solution$
.
So we need more information to find the values of $x$ and $y$.
Let us try to solve the same problem again, but this time with one additional equation.
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How many possible values of $x$ and $y$, do you think, are there that satisfy both these equations simultaneously?
There is exactly one set of such value
.
Now, we can see that when we have two variables, $x$ and $y$ in this case, we need exactly two equations to find a unique solution for the variables.
Similarly if we have three variables, like $a$, $b$ and $c$, we will need
$three$
equations to solve them.
These equations are called $\unicode{0x201C} \text{system of equations}\unicode{0x201D}$ or $\unicode{0x201C}\text{simultaneous equations}\unicode{0x201D}$ because when we find a solution of the variables, the solution satisfies all the given equations simultaneously.
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II. Solving Simultaneous Equations
Before, we start let us understand that both sides of an equation are numbers, and very specifically exactly the same number.
If we have an equation like $3x + 5y = 3y + 12$, which means that whatever values of $x$ and $y$ we take, they should give the same value of $3x + 5y$ and $3y + 12$.
Therefore, we can always perform the same operation on both sides of the equation with the same operand, for example we can multiply both sides by the same number, divided both sides by the same number (except by $0$), add the same number, subtract the same number, take a square of both sides or some other more complex operation. As long as the operation and the operand on both sides are the same, the equality still holds.
When we have two equations of the form:
$A = B$ $...eqn\ (i)$
$C = D$ $...eqn\ (ii)$
and we treat the $L.H.S$ and $R.H.S$ of these two equations as same numbers, then we can see that we can perform all the operations using these two equations as well.
For example, we can say:
adding $eqn\ (ii)$ with $eqn\ (i)$ we get
$A + C =$
$B + D$
or, subtracting $eqn\ (ii)$ from $eqn\ (i)$ we get
$A - C =$
$B - D$
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We can add or subtract any multiple of the given equations like:
$mA \pm nC =$
$mB \pm nD$
or, multiplying both equations we get
$A \times C =$
$B \times D$
or, dividing $eqn\ (i)$ by $eqn\ (ii)$ we get
$A \div C =$
$B \div D$
.
Here we need to be careful, that the divisors $C$ and $D$ are not equal to
$0$
.
Likewise we can perform any other operation using these equations as the operands.
Let us try to solve the following problem:
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Now, let us see how we can use these concepts to solve some simultaneous equations with $2$ variables. As a first step, let us understand the concept of elimination.
Let us try the problem below:
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In this example we eliminated the variable $y$, to come with an equation involving only $x$. Observe that we go different looking equations involving $x$, but on solving both these equations, we get the same value $x$, which is $x = 3$.
Now we can solve for $y$ by substituting the value of $x$ in either of the two equation and then solving for $x$.
Substituting $x = 3$ in the $eqn(i)$ gives us:
$4(3) + 3y = 18$
$\Rightarrow 3y = 6$
$\Rightarrow$
$y = 2$
Substituting the same value of $x$ in $eqn(ii)$ gives us:
$5(3) - 2y = 11$
$\Rightarrow -2y = 11 - 15$
$\Rightarrow -2y = -4$
$\Rightarrow$
$y = 2$
As you can see that the values of $y$ that we get from either equation is
the same
.
Therefore, while selecting which equation to use to find the second variable, you should select the smaller equation or the easy one to solve.
As a final example, let us try the next problem.
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