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Question

Question: Solve the following pair of equations. \(\begin{aligned} & \dfrac{3}{2}x+\dfrac{5}{3}y=7 \\\ ...

Solve the following pair of equations.
32x+53y=7 9x10y=14 \begin{aligned} & \dfrac{3}{2}x+\dfrac{5}{3}y=7 \\\ & 9x-10y=14 \\\ \end{aligned}

Explanation

Solution

For solving these problems, we need to have a complete understanding of simultaneous equations and what is the procedure for solving them. By algebraically solving them, we can easily get the values of x and y respectively.

Complete step-by-step solution:
Simultaneous equations are two equations, each with the same two unknowns and are "simultaneous" because they are solved together. We can find solutions to simultaneous equations algebraically. There are two common methods, algebraically and graphically. Which one we choose might depend on the values involved or it might just be the method one likes the most.
Now, as given in the problem, we need to solve the values of x and y respectively.
32x+53y=7 9x10y=14 \begin{aligned} & \dfrac{3}{2}x+\dfrac{5}{3}y=7 \\\ & 9x-10y=14 \\\ \end{aligned}
With this pair of equations, we will need to multiply both to get a common multiple of either x or y. Now, the coefficient of x in the first equation is 32\dfrac{3}{2} . If we multiply the first equation with 66 , we can get the coefficient of x as 32×6=9\dfrac{3}{2}\times 6=9 which is the same coefficient of x in the second equation,
\left\\{ \dfrac{3}{2}x+\dfrac{5}{3}y=7 \right\\}\times 6=9x+10y=42
9x10y=149x-10y=14
Now subtract the second equation from the first, the terms of x get cancelled out and we are left with the terms of y and the term independent of x and y. We get,
20y=28\Rightarrow 20y=28
Now using division operation, we can get the value of y as 75\dfrac{7}{5} . Substituting the value of y in the second equation, we get that,
9x10(75)=14 9x=28 x=289 \begin{aligned} & \Rightarrow 9x-10\left( \dfrac{7}{5} \right)=14 \\\ & \Rightarrow 9x=28 \\\ & \Rightarrow x=\dfrac{28}{9} \\\ \end{aligned}
Thus, by solving algebraically, we get the values of x and y as 289\dfrac{28}{9} and 75\dfrac{7}{5}.

Note: These types of problems are pretty easy to solve but a slight misjudgement in the calculation can lead to a totally different answer. These simultaneous equations can also be solved graphically by calculating the intersection point of the two lines drawn on a graph paper. After calculating, to be sure about the answer, one should also replace the values of x and y in any of the two equations and check whether the values satisfy the equation.