Solving linear simultaneous equations is an important skill in algebra, especially in Year 12 Math. One great way to do this is by using the substitution method. Let’s break it down step by step.
Look at these two equations:
In the first equation, ( y ) is already expressed using ( x ). This is important because we can plug this expression for ( y ) into the second equation.
Now, let’s take the expression for ( y ) from the first equation and put it into the second equation:
[ 3x + 4(2x + 3) = 10 ]
Here, we replaced ( y ) with ( 2x + 3 ). It might look tricky, but we can solve it together!
Let’s simplify the equation step by step:
First, we multiply the ( 4 ) across the parentheses: [ 3x + 8x + 12 = 10 ]
Now, let’s combine the like terms: [ 11x + 12 = 10 ]
Next, subtract ( 12 ) from both sides: [ 11x = 10 - 12 ] [ 11x = -2 ]
Now, divide both sides by ( 11 ): [ x = \frac{-2}{11} ]
Now that we have ( x ), we need to find ( y ). We can put ( x ) back into the first equation:
[ y = 2\left(\frac{-2}{11}\right) + 3 ]
Let’s calculate this: [ y = \frac{-4}{11} + \frac{33}{11} = \frac{29}{11} ]
So, the solution to the simultaneous equations is:
[ x = \frac{-2}{11}, \quad y = \frac{29}{11} ]
We can write the solution as a pair:
[ \left(\frac{-2}{11}, \frac{29}{11}\right) ]
To recap, the substitution method involves:
This method is especially helpful for solving equations that are easy to work with. With practice, you’ll get better at finding the best way to solve simultaneous equations. Keep practicing, and soon it will feel easy!
Solving linear simultaneous equations is an important skill in algebra, especially in Year 12 Math. One great way to do this is by using the substitution method. Let’s break it down step by step.
Look at these two equations:
In the first equation, ( y ) is already expressed using ( x ). This is important because we can plug this expression for ( y ) into the second equation.
Now, let’s take the expression for ( y ) from the first equation and put it into the second equation:
[ 3x + 4(2x + 3) = 10 ]
Here, we replaced ( y ) with ( 2x + 3 ). It might look tricky, but we can solve it together!
Let’s simplify the equation step by step:
First, we multiply the ( 4 ) across the parentheses: [ 3x + 8x + 12 = 10 ]
Now, let’s combine the like terms: [ 11x + 12 = 10 ]
Next, subtract ( 12 ) from both sides: [ 11x = 10 - 12 ] [ 11x = -2 ]
Now, divide both sides by ( 11 ): [ x = \frac{-2}{11} ]
Now that we have ( x ), we need to find ( y ). We can put ( x ) back into the first equation:
[ y = 2\left(\frac{-2}{11}\right) + 3 ]
Let’s calculate this: [ y = \frac{-4}{11} + \frac{33}{11} = \frac{29}{11} ]
So, the solution to the simultaneous equations is:
[ x = \frac{-2}{11}, \quad y = \frac{29}{11} ]
We can write the solution as a pair:
[ \left(\frac{-2}{11}, \frac{29}{11}\right) ]
To recap, the substitution method involves:
This method is especially helpful for solving equations that are easy to work with. With practice, you’ll get better at finding the best way to solve simultaneous equations. Keep practicing, and soon it will feel easy!