Question

How do you solve the system of equations 3x-6y=24 and -5x+10y=-40?

Answer 1

Hint: This type of problem is based on the concept of linear equations with two variables. Here, we first consider the equation 3x-6y=24. Then, divide the whole equation by 3 and find the value of x. Now, consider the equation -5x+10y=-40 and divide the whole equation by -5. Then, add 2y in both the sides of the equation and find the value of x. equate both the obtained equations with variable y. Do necessary calculations and find the value of y. Then, substitute the value of y in -5x+10y=-40 and obtain the value of x.

Complete step by step answer:
According to the question, we are asked to solve the system of equations 3x-6y=24 and -5x+10y=-40.
We have been given the two equation are
3x-6y=24 ---------(1) and
-5x+10y=-40---------(2)
We first have to consider the equation (1).
Divide the whole equation by 3.
\[\Rightarrow \dfrac{3x-6y}{3}=\dfrac{24}{3}\]
\[\Rightarrow \dfrac{3x}{3}-\dfrac{6y}{3}=\dfrac{24}{3}\]
We can simplify the equation as
\[\dfrac{3x}{3}-\dfrac{3\times 2y}{3}=\dfrac{3\times 8}{3}\]
We find that 3 are common in both the numerator and denominator of LHS and RHS.
On cancelling 3, we get
\[x-2y=8\]
Now add 2y on both the sides of the equation. We get
\[x-2y+2y=8+2y\]
We know that terms with the same magnitude and opposite signs cancel out. We get
x=8+2y -----------(3)
Let us now consider equation (2).
Divide the whole equation by -5.
\[\Rightarrow \dfrac{-5x+10y}{-5}=\dfrac{-40}{-5}\]
\[\Rightarrow \dfrac{-5x}{-5}+\dfrac{10y}{-5}=\dfrac{-40}{-5}\]
We can write the expression as
\[\dfrac{-5x}{-5}+\dfrac{-5\left( -2 \right)y}{-5}=\dfrac{-5\times 8}{-5}\]
We find that -5 are common in both the numerator and denominator of both the LHS and RHS.
On cancelling -5, we get
x-2y=8
Add 2y on both the sides of the equation. We get
x-2y+2y=8+2y
We know that terms with the same magnitude and opposite signs cancel out. We get
x=2y+8 -----------(4)
From equation (3) and (4) we find that both the equations are the same.
Since both the equations are the same, the lines 3x-6y=24 and -5x+10y=-40 coincides with each other.

Therefore, the equations 3x-6y=24 and -5x+10y=-40 have infinite number of solutions.

Note: We can also find the answer in the following method.
Consider the equation (1) and (2).
 The coefficients of x are 3 and -5. Therefore, their ratio is \[\dfrac{3}{-5}=\dfrac{-3}{5}\].
Now, consider the coefficients of y.
Their ratio is \[\dfrac{-6}{10}\]. Here, 2 are divisible by -6 and 10.
Therefore, the ratio of coefficient of y term is \[\dfrac{-3}{5}\].
Now, consider the constants.
The ratio of the constants are \[\dfrac{24}{-40}\].
We know that 24 and -40 is divisible by 4. We can write the ratio as\[\dfrac{6}{-10}\].
 Here, 2 are divisible by 6 and -10.
Therefore, the ratio of coefficient of y term is \[\dfrac{3}{-5}=\dfrac{-3}{5}\].
We find that the ratios of the coefficients of x, y and the constants are the same.
Hence, the given equations coincide with each other.
Therefore, there are an infinite number of solutions for the system of equations.