Simultaneous Equation

In this chapter, we will be discussing the below-mentioned topics in detail:

• Solving Simultaneous Equations using the substitution and elimination method
• Formulate a pair of linear equations in two variables to solve mathematical and real-life problems.
   

Graphical Method

1. Draw the graphs of the given equations.
2. Find the coordinates of the point of intersection.
3. Express coordinates as a solution.
   i.e. \(\begin{align*} x &= \text{__________} \end{align*}\), \(\begin{align*} y &= \text{__________} \end{align*}\).
    

Substitution Method

1. Make x or y the subject of one of the equations.
2. Substitute that equation into the other equation.
3. Solve and present your answers.
    

Elimination Method

1. Make sure the coefficients of one variable are the same.
2. Add or subtract the equations to eliminate one variable.
3. Solve and present your answers.

Let’s understand this with the help of some examples:

Question 1: 

Solve the simultaneous equations using Substitution Method.

\(\begin{align} 2x &= 4 + 3y \\ \\ 3 &= x-y \end{align}\)

Solution:

Using substitution method,

Let us label these equations as Equation (1) and Equation (2),

\(\begin{align} 2x &= 4 + 3y & \text{.......... (1)} \\ \\ 3 &= x-y & \text{.......... (2)} \end{align}\)

Method 1:

From equation (2), make \(x\) the subject.

\(\begin{align} 3+y &=x & \text{.......... (3)} \end{align}\)

Substituting (3) into (1),

\(\begin{align*} 2(3+y) &= 4+3y \\ \\ 6+2y &= 4+3y \\ \\ 2y-3y &= 4-6 \\ \\ -y &=-2 \\ \\ \therefore \qquad\quad y&=2 \end{align*}\)

Substituting \(y=2\) into Equation (3),

\(\begin{align*} 3+2 &=x \\ \\ \therefore\qquad x &=5 \end{align*}\)

Hence, \(x=5\), \(y=2\).

Method 2:

From equation (2), make y the subject.

\(\begin{align} y &= x-3 & \text{.......... (4)} \end{align}\)

Substituting (4) into (1),

\(\begin{align*} 2x&=4+3(x-3)\\\\ 2x&=4+3x-9\\\\ 2x-3x&=-5\\\\ -x&=-5\\\\ \therefore\qquad\quad x&=5 \end{align*}\)

Hence, \(x=5\)

Substituting \(x=5\)  into Equation (3)

\(\begin{align*} y&=5-3\\\\ \therefore\quad y&=2 \end{align*}\)

Hence, \(x=5\), \(y=2\).

Question 2:

Solve the simultaneous equations.

\(\begin{align} 3x+2y &=14 \\ \\ 4y &=13-x \end{align}\)

Solution:

Using substitution method,

Let us label these equations as Equation (1) and Equation (2),

\(\begin{align} 3x+2y &=14 & \text{.......... (1)} \\ \\ 4y &=13- x & \text{.......... (2)} \end{align}\)

From equation (2), make \(x\) the subject.

\(\begin{align} x &=13-4y & \text{.......... (3)} \end{align}\)

Substituting (3) into (1),

\(\begin{align*} 3(13-4y)+2y &=14 \\ \\ 39-12y+2y &=14 \\ \\ -25&=-10y \end{align*}\)

Hence, \(\begin{align} y = 2\frac {1}{2} \end{align}\)

Substituting\(\begin{align} y = 2\frac {1}{2} \end{align}\)into Equation (3),

\(\begin{align*} x&=13-4(2\frac{1}{2}) \\ \\ &=3 \\ \\ \therefore\quad x &=3 \quad \& \quad y = 2\frac {1}{2} \end{align*}\)

Question 3:

Solve the simultaneous equations using Elimination Method.

\(\begin{align} 4x+y &=3 \\ \\ 2x-3y &=12 \end{align}\)

Solution: 

Using elimination method,

Let us label these equations as Equation (1) and Equation (2),

\(\begin{align} 4x+y &=3 & \text{.......... (1)} \\ \\ 2x-3y &=12 & \text{.......... (2)} \end{align}\)

From equation (2), we multiply 2 throughout the equation.

\(\begin{align} 4x-6y &=24 & \text{.......... (3)} \end{align}\)

Taking \(\begin{align} \text{(1) – (2)} \end{align}\),

\(\begin{align*} (4x+y)-(4x-6y)&=3-24\\\\ 7y&=-21\\\\y&=-3 \end{align*}\)

Substituting\(\begin{align} y &=-3 \end{align}\)into Equation (1),

\(\begin{align*} 4x+(-3) &=3 \\ \\ 4x &=6 \\ \\ x &= 1\frac {1}{2} \\ \\ \therefore\qquad\qquad x &= 1\frac{1}{2} \quad \& \quad y=-3 \end{align*}\)