QuestionGiven $x=1$ and $y=-1$ , find two equations for constants $a$ and $b$ from $k(1) + b(-1) = -3$ and $b(1) - a(-1) = 15$ . Then, solve for $a$ and $b$ .

Studdy Solution

STEP 1

Assumptions1. The solution to the simultaneous equations is given as $x=1$ and $y=-1$ . . The simultaneous equations are $kx+by=-3$ and $bx-ay=15$ .
3. $a$ and $b$ are constants.

STEP 2

We can substitute the given values of $x$ and $y$ into the two equations to form two new equations in terms of $a$ and $b$ .
For the first equation, substitute $x=1$ and $y=-1$ $k(1) + b(-1) = -$ For the second equation, substitute $x=1$ and $y=-1$ $b(1) - a(-1) =15$

STEP 3

implify the equations to get two linear equations relating $a$ and $b$ .
For the first equation $k - b = -3$ For the second equation $b + a =15$

STEP 4

From the first equation, we can express $b$ in terms of $k$ $b = k +3$

STEP 5

Substitute $b = k +3$ into the second equation $(k +3) + a =15$

STEP 6

implify the equation to express $a$ in terms of $k$ $a =15 - (k +3)$

STEP 7

implify further to get the equation for $a$ $a =12 - k$

STEP 8

Now we have two equations $b = k +3$ $a =12 - k$ These are the two linear equations relating $a$ and $b$ .

STEP 9

To find the values of $a$ and $b$ , we need to solve these two equations simultaneously.

STEP 10

Substitute $a =12 - k$ into $b = k +3$ $b = (12 - a) +3$

STEP 11

implify the equation to get $b$ in terms of $a$ $b =15 - a$

STEP 12

Now we have two equations $a =12 - k$ $b =15 - a$ We can solve these equations simultaneously to find the values of $a$ and $b$ .

STEP 13

Substitute $b =15 - a$ into $a =12 - k$ $a =12 - (15 - a)$

STEP 14

implify the equation to get $a$ $2a =12 +$

STEP 15

olve for $a$ $a = (12 +15) /2 =13.5$

STEP 16

Substitute $a =13.5$ into $b =15 - a$ to find $b$ $b =15 -13.5 =.5$ So, $a =13.5$ and $b =.5$ .

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QuestionGiven $x=1$ and $y=-1$ , find two equations for constants $a$ and $b$ from $k(1) + b(-1) = -3$ and $b(1) - a(-1) = 15$ . Then, solve for $a$ and $b$ .

Studdy Solution

STEP 1

Assumptions1. The solution to the simultaneous equations is given as $x=1$ and $y=-1$ . . The simultaneous equations are $kx+by=-3$ and $bx-ay=15$ .
3. $a$ and $b$ are constants.

STEP 2

We can substitute the given values of $x$ and $y$ into the two equations to form two new equations in terms of $a$ and $b$ .
For the first equation, substitute $x=1$ and $y=-1$ $k(1) + b(-1) = -$ For the second equation, substitute $x=1$ and $y=-1$ $b(1) - a(-1) =15$

STEP 3

implify the equations to get two linear equations relating $a$ and $b$ .
For the first equation $k - b = -3$ For the second equation $b + a =15$

STEP 4

From the first equation, we can express $b$ in terms of $k$ $b = k +3$

STEP 5

Substitute $b = k +3$ into the second equation $(k +3) + a =15$

STEP 6

implify the equation to express $a$ in terms of $k$ $a =15 - (k +3)$

STEP 7

implify further to get the equation for $a$ $a =12 - k$

STEP 8

Now we have two equations $b = k +3$ $a =12 - k$ These are the two linear equations relating $a$ and $b$ .

STEP 9

To find the values of $a$ and $b$ , we need to solve these two equations simultaneously.

STEP 10

Substitute $a =12 - k$ into $b = k +3$ $b = (12 - a) +3$

STEP 11

implify the equation to get $b$ in terms of $a$ $b =15 - a$

STEP 12

Now we have two equations $a =12 - k$ $b =15 - a$ We can solve these equations simultaneously to find the values of $a$ and $b$ .

STEP 13

Substitute $b =15 - a$ into $a =12 - k$ $a =12 - (15 - a)$

STEP 14

implify the equation to get $a$ $2a =12 +$

STEP 15

olve for $a$ $a = (12 +15) /2 =13.5$

STEP 16

Substitute $a =13.5$ into $b =15 - a$ to find $b$ $b =15 -13.5 =.5$ So, $a =13.5$ and $b =.5$ .

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QuestionGiven x=1x=1x=1 and y=−1y=-1y=−1, find two equations for constants aaa and bbb from k(1)+b(−1)=−3k(1) + b(-1) = -3k(1)+b(−1)=−3 and b(1)−a(−1)=15b(1) - a(-1) = 15b(1)−a(−1)=15. Then, solve for aaa and bbb.

Studdy Solution

STEP 1

Assumptions1. The solution to the simultaneous equations is given as x=1x=1x=1 and y=−1y=-1y=−1. . The simultaneous equations are kx+by=−3kx+by=-3kx+by=−3 and bx−ay=15bx-ay=15bx−ay=15.3. aaa and bbb are constants.

STEP 2

STEP 3

implify the equations to get two linear equations relating aaa and bbb.For the first equationk−b=−3k - b = -3k−b=−3For the second equationb+a=15b + a =15b+a=15

STEP 4

From the first equation, we can express bbb in terms of kkkb=k+3b = k +3b=k+3

STEP 5

Substitute b=k+3b = k +3b=k+3 into the second equation(k+3)+a=15(k +3) + a =15(k+3)+a=15

STEP 6

implify the equation to express aaa in terms of kkka=15−(k+3)a =15 - (k +3)a=15−(k+3)

STEP 7

implify further to get the equation for aaaa=12−ka =12 - ka=12−k

STEP 8

Now we have two equationsb=k+3b = k +3b=k+3a=12−ka =12 - ka=12−kThese are the two linear equations relating aaa and bbb.

STEP 9

To find the values of aaa and bbb, we need to solve these two equations simultaneously.

STEP 10

Substitute a=12−ka =12 - ka=12−k into b=k+3b = k +3b=k+3b=(12−a)+3b = (12 - a) +3b=(12−a)+3

STEP 11

implify the equation to get bbb in terms of aaab=15−ab =15 - ab=15−a

STEP 12

Now we have two equationsa=12−ka =12 - ka=12−kb=15−ab =15 - ab=15−aWe can solve these equations simultaneously to find the values of aaa and bbb.

STEP 13

Substitute b=15−ab =15 - ab=15−a into a=12−ka =12 - ka=12−ka=12−(15−a)a =12 - (15 - a)a=12−(15−a)

STEP 14

implify the equation to get aaa2a=12+2a =12 +2a=12+

STEP 15

olve for aaaa=(12+15)/2=13.5a = (12 +15) /2 =13.5a=(12+15)/2=13.5

STEP 16

Substitute a=13.5a =13.5a=13.5 into b=15−ab =15 - ab=15−a to find bbbb=15−13.5=.5b =15 -13.5 =.5b=15−13.5=.5So, a=13.5a =13.5a=13.5 and b=.5b =.5b=.5.

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QuestionGiven $x=1$ and $y=-1$ , find two equations for constants $a$ and $b$ from $k(1) + b(-1) = -3$ and $b(1) - a(-1) = 15$ . Then, solve for $a$ and $b$ .

Assumptions1. The solution to the simultaneous equations is given as $x=1$ and $y=-1$ . . The simultaneous equations are $kx+by=-3$ and $bx-ay=15$ .
3. $a$ and $b$ are constants.

implify the equations to get two linear equations relating $a$ and $b$ .
For the first equation $k - b = -3$ For the second equation $b + a =15$

From the first equation, we can express $b$ in terms of $k$ $b = k +3$

Substitute $b = k +3$ into the second equation $(k +3) + a =15$

implify the equation to express $a$ in terms of $k$ $a =15 - (k +3)$

implify further to get the equation for $a$ $a =12 - k$

Now we have two equations $b = k +3$ $a =12 - k$ These are the two linear equations relating $a$ and $b$ .

To find the values of $a$ and $b$ , we need to solve these two equations simultaneously.

Substitute $a =12 - k$ into $b = k +3$ $b = (12 - a) +3$

implify the equation to get $b$ in terms of $a$ $b =15 - a$

Now we have two equations $a =12 - k$ $b =15 - a$ We can solve these equations simultaneously to find the values of $a$ and $b$ .

Substitute $b =15 - a$ into $a =12 - k$ $a =12 - (15 - a)$

implify the equation to get $a$ $2a =12 +$

olve for $a$ $a = (12 +15) /2 =13.5$

Substitute $a =13.5$ into $b =15 - a$ to find $b$ $b =15 -13.5 =.5$ So, $a =13.5$ and $b =.5$ .