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
- Draw the graphs of the given equations.
- Find the coordinates of the point of intersection.
- Express coordinates as a solution.
i.e. \(\begin{align*} x &= \text{__________} \end{align*}\), \(\begin{align*} y &= \text{__________} \end{align*}\).
Substitution Method
- Make x or y the subject of one of the equations.
- Substitute that equation into the other equation.
- Solve and present your answers.
Elimination Method
- Make sure the coefficients of one variable are the same.
- Add or subtract the equations to eliminate one variable.
- 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),
\(\begin{align*} 4x+(-3) &=3 \\ \\ 4x &=6 \\ \\ x &= 1\frac {1}{2} \\ \\ \therefore\qquad\qquad x &= 1\frac{1}{2} \quad \& \quad y=-3 \end{align*}\)
Continue Learning | |
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Algebraic Fractions | Direct & Inverse Proportion |
Congruence And Similarity | Factorising Quadratic Expressions |
Further Expansion And Factorisation | Quadratic Equations And Graphs |
Simultaneous Equation |