Solve the following equations using elimination method:
$3x+4y=25$ and $5x-6y=-9$.
Answer
682.8k+ views
Hint: In this question, we will first eliminate one of the variables, by making terms of that variable equal using LCM. And then solve the equations to get required values.
Complete step-by-step answer:
In elimination method, we are to first eliminate one of the variables and then solve the equation for the remaining variable and use the value of that variable to find the value of the first variable.
Now, the given equations are,
$3x+4y=25\cdots \cdots \left( i \right)$
and, $5x-6y=-9\cdots \cdots \left( ii \right)$.
Let us first eliminate $y$. For this, we have to make terms containing $y$ in both the equations equal.
To do this, let us take LCM of coefficients of $y$ from both the equations, which is 4 and 6. So, required LCM will be 12.
Now, to make coefficients of $y$ in both the equation 12, let us first multiply equation $\left( i \right)$ with 3, so we get,
$3\times \left( 3x+4y \right)=3\times 25$
Applying distributive law, we get,
$9x+12y=75\cdots \cdots \left( iii \right)$.
And, now we multiplying equation $\left( ii \right)$ with 2, so we get,
$2\times \left( 5x-6y \right)=2\times \left( -9 \right)$
Applying distributive law, we get,
$\Rightarrow 10x-12y=-18\cdots \cdots \left( iv \right)$
Now, to eliminate $y$, let us add equation $\left( iii \right)$ and $\left( iv \right)$, so we get,
$\begin{align}
& \left( 9x+12y \right)+\left( 10x-12y \right)=75+\left( -18 \right) \\
& \Rightarrow 9x+12y+10x-12y=75-18 \\
& \Rightarrow 9x+10x=75-18 \\
& \Rightarrow 19x=57 \\
\end{align}$
Dividing 19 from both sides of the equation, we get,
$x=\dfrac{57}{19}=3$.
Now, putting this value of $x$ in equation $\left( i \right)$, we get,
$\begin{align}
& 3x+4y=25 \\
& \Rightarrow 3\times 3+4y=25 \\
& \Rightarrow 9+4y=25 \\
\end{align}$
Subtracting 9 from both sides of the above equation, we get,
$\begin{align}
& 4y=25-9 \\
& \Rightarrow 4y=16 \\
\end{align}$
Dividing 4 from both sides of the equation, we get,
$y=\dfrac{16}{4}=4$.
Hence, in given equations, the value of $x$ is 3 and value of $y$ is 4.
Note: In this type of question, you can eliminate any one of the variables, but choose that variable for which equations will be multiplied by a smaller number so as to keep calculation easier.
Complete step-by-step answer:
In elimination method, we are to first eliminate one of the variables and then solve the equation for the remaining variable and use the value of that variable to find the value of the first variable.
Now, the given equations are,
$3x+4y=25\cdots \cdots \left( i \right)$
and, $5x-6y=-9\cdots \cdots \left( ii \right)$.
Let us first eliminate $y$. For this, we have to make terms containing $y$ in both the equations equal.
To do this, let us take LCM of coefficients of $y$ from both the equations, which is 4 and 6. So, required LCM will be 12.
Now, to make coefficients of $y$ in both the equation 12, let us first multiply equation $\left( i \right)$ with 3, so we get,
$3\times \left( 3x+4y \right)=3\times 25$
Applying distributive law, we get,
$9x+12y=75\cdots \cdots \left( iii \right)$.
And, now we multiplying equation $\left( ii \right)$ with 2, so we get,
$2\times \left( 5x-6y \right)=2\times \left( -9 \right)$
Applying distributive law, we get,
$\Rightarrow 10x-12y=-18\cdots \cdots \left( iv \right)$
Now, to eliminate $y$, let us add equation $\left( iii \right)$ and $\left( iv \right)$, so we get,
$\begin{align}
& \left( 9x+12y \right)+\left( 10x-12y \right)=75+\left( -18 \right) \\
& \Rightarrow 9x+12y+10x-12y=75-18 \\
& \Rightarrow 9x+10x=75-18 \\
& \Rightarrow 19x=57 \\
\end{align}$
Dividing 19 from both sides of the equation, we get,
$x=\dfrac{57}{19}=3$.
Now, putting this value of $x$ in equation $\left( i \right)$, we get,
$\begin{align}
& 3x+4y=25 \\
& \Rightarrow 3\times 3+4y=25 \\
& \Rightarrow 9+4y=25 \\
\end{align}$
Subtracting 9 from both sides of the above equation, we get,
$\begin{align}
& 4y=25-9 \\
& \Rightarrow 4y=16 \\
\end{align}$
Dividing 4 from both sides of the equation, we get,
$y=\dfrac{16}{4}=4$.
Hence, in given equations, the value of $x$ is 3 and value of $y$ is 4.
Note: In this type of question, you can eliminate any one of the variables, but choose that variable for which equations will be multiplied by a smaller number so as to keep calculation easier.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

