Make y the subject of x = y/(y-z)
$begingroup$
I'm struggling with this GCSE question, but I think I'm just being silly. I've removed the fraction, making it:
x(y-z) = y
And then tried removing the brackets, making it:
xy-xz = y
But I'm not sure how to get all of the y terms onto one side of the equation in this situation, or whether I've gone about this the wrong way.
Any help would be appreciated! Retaking my GCSE and I'm extremely rusty!
Thanks,
Jay
recreational-mathematics
asked Jul 15 '16 at 16:55
JayJay
62
$endgroup$
add a comment |
$begingroup$
I'm struggling with this GCSE question, but I think I'm just being silly. I've removed the fraction, making it:
x(y-z) = y
And then tried removing the brackets, making it:
xy-xz = y
But I'm not sure how to get all of the y terms onto one side of the equation in this situation, or whether I've gone about this the wrong way.
Any help would be appreciated! Retaking my GCSE and I'm extremely rusty!
Thanks,
Jay
recreational-mathematics
asked Jul 15 '16 at 16:55
JayJay
62
$endgroup$
$begingroup$
For your next step, say $xy-y=xz$
$endgroup$
– turkeyhundt
Jul 15 '16 at 17:00
1
$begingroup$
The key to getting all the $y$ terms on one side is to ... get all the y terms on one side. So move the $xy$ term to the other side, as it has a y in it, then factor that and get that $y = frac{xz}{x-1}$.
$endgroup$
– Klint Qinami
Jul 15 '16 at 17:06
$begingroup$
Ah, I understand now. Thanks for your time! - Jay
$endgroup$
– Jay
Jul 15 '16 at 17:26
$begingroup$
What is a "subject" in mathematics?
$endgroup$
– Christian Blatter
Jul 15 '16 at 17:58
add a comment |
$begingroup$
I'm struggling with this GCSE question, but I think I'm just being silly. I've removed the fraction, making it:
x(y-z) = y
And then tried removing the brackets, making it:
xy-xz = y
But I'm not sure how to get all of the y terms onto one side of the equation in this situation, or whether I've gone about this the wrong way.
Any help would be appreciated! Retaking my GCSE and I'm extremely rusty!
Thanks,
Jay
recreational-mathematics
asked Jul 15 '16 at 16:55
JayJay
62
$endgroup$
I'm struggling with this GCSE question, but I think I'm just being silly. I've removed the fraction, making it:
x(y-z) = y
And then tried removing the brackets, making it:
xy-xz = y
But I'm not sure how to get all of the y terms onto one side of the equation in this situation, or whether I've gone about this the wrong way.
Any help would be appreciated! Retaking my GCSE and I'm extremely rusty!
Thanks,
Jay
recreational-mathematics
recreational-mathematics
asked Jul 15 '16 at 16:55
JayJay
62
asked Jul 15 '16 at 16:55
JayJay
62
asked Jul 15 '16 at 16:55
JayJay
62
asked Jul 15 '16 at 16:55
JayJay
62
asked Jul 15 '16 at 16:55
JayJay
62
62
$begingroup$
For your next step, say $xy-y=xz$
$endgroup$
– turkeyhundt
Jul 15 '16 at 17:00
1
$begingroup$
The key to getting all the $y$ terms on one side is to ... get all the y terms on one side. So move the $xy$ term to the other side, as it has a y in it, then factor that and get that $y = frac{xz}{x-1}$.
$endgroup$
– Klint Qinami
Jul 15 '16 at 17:06
$begingroup$
Ah, I understand now. Thanks for your time! - Jay
$endgroup$
– Jay
Jul 15 '16 at 17:26
$begingroup$
What is a "subject" in mathematics?
$endgroup$
– Christian Blatter
Jul 15 '16 at 17:58
add a comment |
$begingroup$
For your next step, say $xy-y=xz$
$endgroup$
– turkeyhundt
Jul 15 '16 at 17:00
1
$begingroup$
The key to getting all the $y$ terms on one side is to ... get all the y terms on one side. So move the $xy$ term to the other side, as it has a y in it, then factor that and get that $y = frac{xz}{x-1}$.
$endgroup$
– Klint Qinami
Jul 15 '16 at 17:06
$begingroup$
Ah, I understand now. Thanks for your time! - Jay
$endgroup$
– Jay
Jul 15 '16 at 17:26
$begingroup$
What is a "subject" in mathematics?
$endgroup$
– Christian Blatter
Jul 15 '16 at 17:58
$begingroup$
For your next step, say $xy-y=xz$
$endgroup$
– turkeyhundt
Jul 15 '16 at 17:00
$begingroup$
For your next step, say $xy-y=xz$
$endgroup$
– turkeyhundt
Jul 15 '16 at 17:00
1
1
$begingroup$
The key to getting all the $y$ terms on one side is to ... get all the y terms on one side. So move the $xy$ term to the other side, as it has a y in it, then factor that and get that $y = frac{xz}{x-1}$.
$endgroup$
– Klint Qinami
Jul 15 '16 at 17:06
$begingroup$
The key to getting all the $y$ terms on one side is to ... get all the y terms on one side. So move the $xy$ term to the other side, as it has a y in it, then factor that and get that $y = frac{xz}{x-1}$.
$endgroup$
– Klint Qinami
Jul 15 '16 at 17:06
$begingroup$
Ah, I understand now. Thanks for your time! - Jay
$endgroup$
– Jay
Jul 15 '16 at 17:26
$begingroup$
Ah, I understand now. Thanks for your time! - Jay
$endgroup$
– Jay
Jul 15 '16 at 17:26
$begingroup$
What is a "subject" in mathematics?
$endgroup$
– Christian Blatter
Jul 15 '16 at 17:58
$begingroup$
What is a "subject" in mathematics?
$endgroup$
– Christian Blatter
Jul 15 '16 at 17:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There are four steps to solving algebraic equations which are linear in one of the variables:
SIMPLIFY-SORT-FACTOR-DIVIDE
Example: Solve for $x$
begin{equation}
dfrac{1-ax}{2}=dfrac{frac{1}{x}-1}{frac{3}{x}}
end{equation}
(1) Simplify:
(a) Remove all fractions by reducing any complex fractions then multiplying both sides of the equation by the least common denominator of remaining fractions on either side.
begin{equation}
dfrac{1-ax}{2}=dfrac{frac{1}{x}-1}{frac{3}{x}}cdotdfrac{x}{x}
end{equation}
begin{equation}
dfrac{1-ax}{2}=dfrac{1-x}{3}
end{equation}
begin{equation}
6cdotdfrac{1-ax}{2}=6cdotdfrac{1-x}{3}
end{equation}
begin{equation}
3(1-ax)=2(1-x)
end{equation}
(b) Remove all parentheses using the distributive property.
begin{equation}
3-3ax=2-2x
end{equation}
(c) Combine all like terms appearing on the same side of the equation.
Not applicable in this example.
(2) Sort:
Add or subtract terms to both sides of the equation in order that all terms containing the variable to be solved for are on one side of the equation while all the remaining terms are on the other side.
begin{equation}
2x-3ax=2-3
end{equation}
begin{equation}
2x-3ax=-1
end{equation}
(3) Factor:
Factor out the variable to be solved for from the terms containing it as a factor.
begin{equation}
(2-3a)x=-1
end{equation}
(4) Divide:
Divide both sides by the coefficient of the factor of the variable to be solved for.
begin{equation}
x=dfrac{-1}{2-3a}=dfrac{1}{3a-2}
end{equation}
Note that at the conclusion of the SIMPLIFY step you may solve for any of the other variables for which the equation is linear. As an exercise you might go back to the simplified equation after step 1b and solve the equation for the variable $a$ in terms of the variable $x$.
answered Jul 15 '16 at 17:31
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are four steps to solving algebraic equations which are linear in one of the variables:
SIMPLIFY-SORT-FACTOR-DIVIDE
Example: Solve for $x$
begin{equation}
dfrac{1-ax}{2}=dfrac{frac{1}{x}-1}{frac{3}{x}}
end{equation}
(1) Simplify:
(a) Remove all fractions by reducing any complex fractions then multiplying both sides of the equation by the least common denominator of remaining fractions on either side.
begin{equation}
dfrac{1-ax}{2}=dfrac{frac{1}{x}-1}{frac{3}{x}}cdotdfrac{x}{x}
end{equation}
begin{equation}
dfrac{1-ax}{2}=dfrac{1-x}{3}
end{equation}
begin{equation}
6cdotdfrac{1-ax}{2}=6cdotdfrac{1-x}{3}
end{equation}
begin{equation}
3(1-ax)=2(1-x)
end{equation}
(b) Remove all parentheses using the distributive property.
begin{equation}
3-3ax=2-2x
end{equation}
(c) Combine all like terms appearing on the same side of the equation.
Not applicable in this example.
(2) Sort:
Add or subtract terms to both sides of the equation in order that all terms containing the variable to be solved for are on one side of the equation while all the remaining terms are on the other side.
begin{equation}
2x-3ax=2-3
end{equation}
begin{equation}
2x-3ax=-1
end{equation}
(3) Factor:
Factor out the variable to be solved for from the terms containing it as a factor.
begin{equation}
(2-3a)x=-1
end{equation}
(4) Divide:
Divide both sides by the coefficient of the factor of the variable to be solved for.
begin{equation}
x=dfrac{-1}{2-3a}=dfrac{1}{3a-2}
end{equation}
Note that at the conclusion of the SIMPLIFY step you may solve for any of the other variables for which the equation is linear. As an exercise you might go back to the simplified equation after step 1b and solve the equation for the variable $a$ in terms of the variable $x$.
answered Jul 15 '16 at 17:31
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
add a comment |
$begingroup$
There are four steps to solving algebraic equations which are linear in one of the variables:
SIMPLIFY-SORT-FACTOR-DIVIDE
Example: Solve for $x$
begin{equation}
dfrac{1-ax}{2}=dfrac{frac{1}{x}-1}{frac{3}{x}}
end{equation}
(1) Simplify:
(a) Remove all fractions by reducing any complex fractions then multiplying both sides of the equation by the least common denominator of remaining fractions on either side.
begin{equation}
dfrac{1-ax}{2}=dfrac{frac{1}{x}-1}{frac{3}{x}}cdotdfrac{x}{x}
end{equation}
begin{equation}
dfrac{1-ax}{2}=dfrac{1-x}{3}
end{equation}
begin{equation}
6cdotdfrac{1-ax}{2}=6cdotdfrac{1-x}{3}
end{equation}
begin{equation}
3(1-ax)=2(1-x)
end{equation}
(b) Remove all parentheses using the distributive property.
begin{equation}
3-3ax=2-2x
end{equation}
(c) Combine all like terms appearing on the same side of the equation.
Not applicable in this example.
(2) Sort:
Add or subtract terms to both sides of the equation in order that all terms containing the variable to be solved for are on one side of the equation while all the remaining terms are on the other side.
begin{equation}
2x-3ax=2-3
end{equation}
begin{equation}
2x-3ax=-1
end{equation}
(3) Factor:
Factor out the variable to be solved for from the terms containing it as a factor.
begin{equation}
(2-3a)x=-1
end{equation}
(4) Divide:
Divide both sides by the coefficient of the factor of the variable to be solved for.
begin{equation}
x=dfrac{-1}{2-3a}=dfrac{1}{3a-2}
end{equation}
Note that at the conclusion of the SIMPLIFY step you may solve for any of the other variables for which the equation is linear. As an exercise you might go back to the simplified equation after step 1b and solve the equation for the variable $a$ in terms of the variable $x$.
answered Jul 15 '16 at 17:31
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
add a comment |
$begingroup$
There are four steps to solving algebraic equations which are linear in one of the variables:
SIMPLIFY-SORT-FACTOR-DIVIDE
Example: Solve for $x$
begin{equation}
dfrac{1-ax}{2}=dfrac{frac{1}{x}-1}{frac{3}{x}}
end{equation}
(1) Simplify:
(a) Remove all fractions by reducing any complex fractions then multiplying both sides of the equation by the least common denominator of remaining fractions on either side.
begin{equation}
dfrac{1-ax}{2}=dfrac{frac{1}{x}-1}{frac{3}{x}}cdotdfrac{x}{x}
end{equation}
begin{equation}
dfrac{1-ax}{2}=dfrac{1-x}{3}
end{equation}
begin{equation}
6cdotdfrac{1-ax}{2}=6cdotdfrac{1-x}{3}
end{equation}
begin{equation}
3(1-ax)=2(1-x)
end{equation}
(b) Remove all parentheses using the distributive property.
begin{equation}
3-3ax=2-2x
end{equation}
(c) Combine all like terms appearing on the same side of the equation.
Not applicable in this example.
(2) Sort:
Add or subtract terms to both sides of the equation in order that all terms containing the variable to be solved for are on one side of the equation while all the remaining terms are on the other side.
begin{equation}
2x-3ax=2-3
end{equation}
begin{equation}
2x-3ax=-1
end{equation}
(3) Factor:
Factor out the variable to be solved for from the terms containing it as a factor.
begin{equation}
(2-3a)x=-1
end{equation}
(4) Divide:
Divide both sides by the coefficient of the factor of the variable to be solved for.
begin{equation}
x=dfrac{-1}{2-3a}=dfrac{1}{3a-2}
end{equation}
Note that at the conclusion of the SIMPLIFY step you may solve for any of the other variables for which the equation is linear. As an exercise you might go back to the simplified equation after step 1b and solve the equation for the variable $a$ in terms of the variable $x$.
answered Jul 15 '16 at 17:31
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
There are four steps to solving algebraic equations which are linear in one of the variables:
SIMPLIFY-SORT-FACTOR-DIVIDE
Example: Solve for $x$
begin{equation}
dfrac{1-ax}{2}=dfrac{frac{1}{x}-1}{frac{3}{x}}
end{equation}
(1) Simplify:
(a) Remove all fractions by reducing any complex fractions then multiplying both sides of the equation by the least common denominator of remaining fractions on either side.
begin{equation}
dfrac{1-ax}{2}=dfrac{frac{1}{x}-1}{frac{3}{x}}cdotdfrac{x}{x}
end{equation}
begin{equation}
dfrac{1-ax}{2}=dfrac{1-x}{3}
end{equation}
begin{equation}
6cdotdfrac{1-ax}{2}=6cdotdfrac{1-x}{3}
end{equation}
begin{equation}
3(1-ax)=2(1-x)
end{equation}
(b) Remove all parentheses using the distributive property.
begin{equation}
3-3ax=2-2x
end{equation}
(c) Combine all like terms appearing on the same side of the equation.
Not applicable in this example.
(2) Sort:
Add or subtract terms to both sides of the equation in order that all terms containing the variable to be solved for are on one side of the equation while all the remaining terms are on the other side.
begin{equation}
2x-3ax=2-3
end{equation}
begin{equation}
2x-3ax=-1
end{equation}
(3) Factor:
Factor out the variable to be solved for from the terms containing it as a factor.
begin{equation}
(2-3a)x=-1
end{equation}
(4) Divide:
Divide both sides by the coefficient of the factor of the variable to be solved for.
begin{equation}
x=dfrac{-1}{2-3a}=dfrac{1}{3a-2}
end{equation}
Note that at the conclusion of the SIMPLIFY step you may solve for any of the other variables for which the equation is linear. As an exercise you might go back to the simplified equation after step 1b and solve the equation for the variable $a$ in terms of the variable $x$.
answered Jul 15 '16 at 17:31
John Wayland BalesJohn Wayland Bales
15.1k21238
answered Jul 15 '16 at 17:31
John Wayland BalesJohn Wayland Bales
15.1k21238
answered Jul 15 '16 at 17:31
John Wayland BalesJohn Wayland Bales
15.1k21238
answered Jul 15 '16 at 17:31
John Wayland BalesJohn Wayland Bales
15.1k21238
15.1k21238
add a comment |
add a comment |
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$begingroup$
For your next step, say $xy-y=xz$
$endgroup$
– turkeyhundt
Jul 15 '16 at 17:00
1
$begingroup$
The key to getting all the $y$ terms on one side is to ... get all the y terms on one side. So move the $xy$ term to the other side, as it has a y in it, then factor that and get that $y = frac{xz}{x-1}$.
$endgroup$
– Klint Qinami
Jul 15 '16 at 17:06
$begingroup$
Ah, I understand now. Thanks for your time! - Jay
$endgroup$
– Jay
Jul 15 '16 at 17:26
$begingroup$
What is a "subject" in mathematics?
$endgroup$
– Christian Blatter
Jul 15 '16 at 17:58