
How do you solve the system of equations $2x-3y=2$ and $3x+5y=-1$ using elimination?
Answer
538.2k+ views
Hint: We will multiply any one of the equations or both equations with a scalar so that the coefficient of any one variable becomes the same. Then using addition or subtraction, we will eliminate that variable to find the value of the other variable. Then we will apply that value in any one of the equations to get the value of the second variable which we had eliminated before.
Complete step by step solution:
Let us consider the given system of equations, $2x-3y=12.......\left( 1 \right)$ and $3x+5y=-1.......\left( 2 \right)$
We are asked to solve the system of equations using elimination.
So, we need to make the coefficients of any one variable in both equations the same. For that, we need to multiply the equations with scalars.
Let us multiply the equation $\left( 1 \right)$ with $3$ and the equation $\left( 2 \right)$ with $2.$ Then the coefficient of the variable $x$ will become $6$ in both equations.
Now, we will get the equation $\left( 1 \right)\times 3$ as $6x-9y=36.......\left( 3 \right)$
And the equation $\left( 2 \right)\times 2$ will give us $6x+10y=-2.......\left( 4 \right)$
Let us subtract the equation$\left( 3 \right)$ from the equation $\left( 4 \right)$ so that we can eliminate the variable $x$ to get $19y=-38.$
From this, when we transpose $19,$ we will get the value of $y$ as $y=\dfrac{-38}{19}=-2.$
Now, we are going to substitute this value in the equation $\left( 1 \right)$ to get the value of $x.$
We will get $2x-3\times \left( -2 \right)=12.$
From this, we will get $2x+6=12.$
So, when we transpose the constant terms, we will get $2x=12-6=6.$
And we will get $x=\dfrac{6}{2}=3.$
Hence the solution of the system is $x=3,y=-2.$
Note: In this method, we can eliminate the variable $y$ instead of the variable $x.$ We just need to multiply the equation $\left( 1 \right)$ with $5$ and the equation $\left( 2 \right)$ with $3.$ Similarly, we can apply the obtained value in the equation $\left( 2 \right).$ The result will be the same.
Complete step by step solution:
Let us consider the given system of equations, $2x-3y=12.......\left( 1 \right)$ and $3x+5y=-1.......\left( 2 \right)$
We are asked to solve the system of equations using elimination.
So, we need to make the coefficients of any one variable in both equations the same. For that, we need to multiply the equations with scalars.
Let us multiply the equation $\left( 1 \right)$ with $3$ and the equation $\left( 2 \right)$ with $2.$ Then the coefficient of the variable $x$ will become $6$ in both equations.
Now, we will get the equation $\left( 1 \right)\times 3$ as $6x-9y=36.......\left( 3 \right)$
And the equation $\left( 2 \right)\times 2$ will give us $6x+10y=-2.......\left( 4 \right)$
Let us subtract the equation$\left( 3 \right)$ from the equation $\left( 4 \right)$ so that we can eliminate the variable $x$ to get $19y=-38.$
From this, when we transpose $19,$ we will get the value of $y$ as $y=\dfrac{-38}{19}=-2.$
Now, we are going to substitute this value in the equation $\left( 1 \right)$ to get the value of $x.$
We will get $2x-3\times \left( -2 \right)=12.$
From this, we will get $2x+6=12.$
So, when we transpose the constant terms, we will get $2x=12-6=6.$
And we will get $x=\dfrac{6}{2}=3.$
Hence the solution of the system is $x=3,y=-2.$
Note: In this method, we can eliminate the variable $y$ instead of the variable $x.$ We just need to multiply the equation $\left( 1 \right)$ with $5$ and the equation $\left( 2 \right)$ with $3.$ Similarly, we can apply the obtained value in the equation $\left( 2 \right).$ The result will be the same.
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