How do you solve the system 5a+3b=17 and 4a-5b=21?
To solve a system of linear equations that are both in standard form, the most efficient method to use would be the "elimination method".
To carry out elimination, you need to alter one (or both) of the equations so as to have one of the variables (and their coefficient) match the same one in another equation.
In your example, you have:
Say I want to solve for b, this means I need to eliminate a. To do this, I need to multiply eq. 1 by 4 and eq. 2 by 5 like so:
This results in:
Now, if the variable you are trying to eliminate has the same sign in both equations, you subtract the equations. If the signs are opposite, you add them. To remember this, think OASS (opposite, add, same, subtract). In this case, the signs are the same so you want to subtract the equations, resulting in:
Now that you have b, you can substitute that back into one of the original equations to solve for a. I'll sub it into eq. 1:
Solving for a:
Therefore, your solution set (or point of intersection) is (4, -1).
Hopefully, this was all clear and I hope you found it helpful! :)
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To solve the system of equations (5a + 3b = 17) and (4a - 5b = 21), you can use the method of substitution or elimination.
Using the substitution method:
- Solve one of the equations for one variable (let's solve the first equation for (a)): (a = \frac{17 - 3b}{5}).
- Substitute the expression for (a) into the other equation: (4\left(\frac{17 - 3b}{5}\right) - 5b = 21).
- Solve for (b).
- Once you find the value of (b), substitute it back into one of the original equations to solve for (a).
Using the elimination method:
- Multiply both equations by suitable constants to make the coefficients of one of the variables the same (for example, if you multiply the first equation by 4 and the second equation by 5, you'll get (20a + 12b = 68) and (20a - 25b = 105)).
- Subtract one equation from the other to eliminate one of the variables.
- Solve for the remaining variable.
- Substitute the value of the variable found back into one of the original equations to solve for the other variable.
Either method will give you the values of (a) and (b) that satisfy both equations.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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