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How do you solve 10a plus 5b? - Answers

You can't just solve one side of the equation. You always have to have both sides of the equal sign in order to manipulate the terms. If you have an equation like 10a + 5b=0, you can now solve for one of the terms (a or b), for example, b=10a/5 and a=-5b/10. Once you have solved for one term, you can plug it back into the equation and solve for real values. In this particular equation, a=-1/2b so by putting any value in for b (say 2), you get a definite value for a (-1) but there are an infinite number of such values.



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How do you solve 10a plus 5b? - Answers

https://math.answers.com/math-and-arithmetic/How_do_you_solve_10a_plus_5b

You can't just solve one side of the equation. You always have to have both sides of the equal sign in order to manipulate the terms. If you have an equation like 10a + 5b=0, you can now solve for one of the terms (a or b), for example, b=10a/5 and a=-5b/10. Once you have solved for one term, you can plug it back into the equation and solve for real values. In this particular equation, a=-1/2b so by putting any value in for b (say 2), you get a definite value for a (-1) but there are an infinite number of such values.



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https://math.answers.com/math-and-arithmetic/How_do_you_solve_10a_plus_5b

How do you solve 10a plus 5b? - Answers

You can't just solve one side of the equation. You always have to have both sides of the equal sign in order to manipulate the terms. If you have an equation like 10a + 5b=0, you can now solve for one of the terms (a or b), for example, b=10a/5 and a=-5b/10. Once you have solved for one term, you can plug it back into the equation and solve for real values. In this particular equation, a=-1/2b so by putting any value in for b (say 2), you get a definite value for a (-1) but there are an infinite number of such values.

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      You can't just solve one side of the equation. You always have to have both sides of the equal sign in order to manipulate the terms. If you have an equation like 10a + 5b=0, you can now solve for one of the terms (a or b), for example, b=10a/5 and a=-5b/10. Once you have solved for one term, you can plug it back into the equation and solve for real values. In this particular equation, a=-1/2b so by putting any value in for b (say 2), you get a definite value for a (-1) but there are an infinite number of such values.

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