QuestionSolve the system of equations: $x - \frac{y}{7} = -1$ and $-\frac{x}{2} + \frac{y}{16} = \frac{1}{4}$ .

Studdy Solution

STEP 1

Assumptions1. We are given a system of two linear equations $\left\{\begin{array}{r}x-\frac{y}{7}=-1 \\ -\frac{x}{}+\frac{y}{16}=\frac{1}{4}\end{array}\right.$ . We need to find the values of $x$ and $y$ that satisfy both equations.

STEP 2

Let's start by isolating $x$ in the first equation. We can do this by adding $\frac{y}{7}$ to both sides of the equation.
$x = \frac{y}{7} -1$

STEP 3

Now, let's substitute $x$ from the first equation into the second equation. This will give us an equation with only $y$ .
$-\frac{1}{2}\left(\frac{y}{7} -1\right) + \frac{y}{16} = \frac{1}{}$

STEP 4

Next, distribute the $-\frac{1}{2}$ inside the parentheses.
$-\frac{y}{14} + \frac{1}{2} + \frac{y}{16} = \frac{1}{4}$

STEP 5

Combine like terms on the left side of the equation.
$\left(-\frac{1}{14} + \frac{1}{16}\right)y + \frac{1}{2} = \frac{1}{4}$

STEP 6

Subtract $\frac{1}{2}$ from both sides of the equation.
$\left(-\frac{1}{14} + \frac{1}{16}\right)y = \frac{1}{4} - \frac{1}{2}$

STEP 7

implify the right side of the equation.
$\left(-\frac{1}{14} + \frac{1}{16}\right)y = -\frac{1}{4}$

STEP 8

implify the coefficient of $y$ on the left side of the equation.
$\left(-\frac{16}{224} + \frac{14}{224}\right)y = -\frac{1}{4}$ $\left(-\frac{2}{224}\right)y = -\frac{1}{4}$ $-\frac{1}{112}y = -\frac{1}{4}$

STEP 9

Multiply both sides of the equation by $-112$ to solve for $y$ .
$y = -\frac{}{4} \times -112$

STEP 10

Calculate the value of $y$ .
$y =28$

STEP 11

Now that we have the value of $y$ , we can substitute it into the first equation to solve for $x$ .
$x = \frac{28}{7} -$

STEP 12

Calculate the value of $x$ .
$x =4 -$ $x =$ The solution to the system of equations is $x =$ and $y =28$ .

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QuestionSolve the system of equations: $x - \frac{y}{7} = -1$ and $-\frac{x}{2} + \frac{y}{16} = \frac{1}{4}$ .

Studdy Solution

STEP 1

Assumptions1. We are given a system of two linear equations $\left\{\begin{array}{r}x-\frac{y}{7}=-1 \\ -\frac{x}{}+\frac{y}{16}=\frac{1}{4}\end{array}\right.$ . We need to find the values of $x$ and $y$ that satisfy both equations.

STEP 2

Let's start by isolating $x$ in the first equation. We can do this by adding $\frac{y}{7}$ to both sides of the equation.
$x = \frac{y}{7} -1$

STEP 3

Now, let's substitute $x$ from the first equation into the second equation. This will give us an equation with only $y$ .
$-\frac{1}{2}\left(\frac{y}{7} -1\right) + \frac{y}{16} = \frac{1}{}$

STEP 4

Next, distribute the $-\frac{1}{2}$ inside the parentheses.
$-\frac{y}{14} + \frac{1}{2} + \frac{y}{16} = \frac{1}{4}$

STEP 5

Combine like terms on the left side of the equation.
$\left(-\frac{1}{14} + \frac{1}{16}\right)y + \frac{1}{2} = \frac{1}{4}$

STEP 6

Subtract $\frac{1}{2}$ from both sides of the equation.
$\left(-\frac{1}{14} + \frac{1}{16}\right)y = \frac{1}{4} - \frac{1}{2}$

STEP 7

implify the right side of the equation.
$\left(-\frac{1}{14} + \frac{1}{16}\right)y = -\frac{1}{4}$

STEP 8

implify the coefficient of $y$ on the left side of the equation.
$\left(-\frac{16}{224} + \frac{14}{224}\right)y = -\frac{1}{4}$ $\left(-\frac{2}{224}\right)y = -\frac{1}{4}$ $-\frac{1}{112}y = -\frac{1}{4}$

STEP 9

Multiply both sides of the equation by $-112$ to solve for $y$ .
$y = -\frac{}{4} \times -112$

STEP 10

Calculate the value of $y$ .
$y =28$

STEP 11

Now that we have the value of $y$ , we can substitute it into the first equation to solve for $x$ .
$x = \frac{28}{7} -$

STEP 12

Calculate the value of $x$ .
$x =4 -$ $x =$ The solution to the system of equations is $x =$ and $y =28$ .

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QuestionSolve the system of equations: x−y7=−1x - \frac{y}{7} = -1x−7y​=−1 and −x2+y16=14-\frac{x}{2} + \frac{y}{16} = \frac{1}{4}−2x​+16y​=41​.

Studdy Solution

STEP 1

Assumptions1. We are given a system of two linear equations {x−y7=−1−x+y16=14\left\{\begin{array}{r}x-\frac{y}{7}=-1 \\ -\frac{x}{}+\frac{y}{16}=\frac{1}{4}\end{array}\right.{x−7y​=−1−x​+16y​=41​​ . We need to find the values of xxx and yyy that satisfy both equations.

STEP 2

Let's start by isolating xxx in the first equation. We can do this by adding y7\frac{y}{7}7y​ to both sides of the equation.x=y7−1x = \frac{y}{7} -1x=7y​−1

STEP 3

Now, let's substitute xxx from the first equation into the second equation. This will give us an equation with only yyy.−12(y7−1)+y16=1-\frac{1}{2}\left(\frac{y}{7} -1\right) + \frac{y}{16} = \frac{1}{}−21​(7y​−1)+16y​=1​

STEP 4

Next, distribute the −12-\frac{1}{2}−21​ inside the parentheses.−y14+12+y16=14-\frac{y}{14} + \frac{1}{2} + \frac{y}{16} = \frac{1}{4}−14y​+21​+16y​=41​

STEP 5

Combine like terms on the left side of the equation.(−114+116)y+12=14\left(-\frac{1}{14} + \frac{1}{16}\right)y + \frac{1}{2} = \frac{1}{4}(−141​+161​)y+21​=41​

STEP 6

Subtract 12\frac{1}{2}21​ from both sides of the equation.(−114+116)y=14−12\left(-\frac{1}{14} + \frac{1}{16}\right)y = \frac{1}{4} - \frac{1}{2}(−141​+161​)y=41​−21​

STEP 7

implify the right side of the equation.(−114+116)y=−14\left(-\frac{1}{14} + \frac{1}{16}\right)y = -\frac{1}{4}(−141​+161​)y=−41​

STEP 8

STEP 9

Multiply both sides of the equation by −112-112−112 to solve for yyy.y=−4×−112y = -\frac{}{4} \times -112y=−4​×−112

STEP 10

Calculate the value of yyy.y=28y =28y=28

STEP 11

Now that we have the value of yyy, we can substitute it into the first equation to solve for xxx.x=287−x = \frac{28}{7} -x=728​−

STEP 12

Calculate the value of xxx.x=4−x =4 -x=4−x=x =x=The solution to the system of equations is x=x =x= and y=28y =28y=28.

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QuestionSolve the system of equations: $x - \frac{y}{7} = -1$ and $-\frac{x}{2} + \frac{y}{16} = \frac{1}{4}$ .

Assumptions1. We are given a system of two linear equations $\left\{\begin{array}{r}x-\frac{y}{7}=-1 \\ -\frac{x}{}+\frac{y}{16}=\frac{1}{4}\end{array}\right.$ . We need to find the values of $x$ and $y$ that satisfy both equations.

Let's start by isolating $x$ in the first equation. We can do this by adding $\frac{y}{7}$ to both sides of the equation.
$x = \frac{y}{7} -1$

Now, let's substitute $x$ from the first equation into the second equation. This will give us an equation with only $y$ .
$-\frac{1}{2}\left(\frac{y}{7} -1\right) + \frac{y}{16} = \frac{1}{}$

Next, distribute the $-\frac{1}{2}$ inside the parentheses.
$-\frac{y}{14} + \frac{1}{2} + \frac{y}{16} = \frac{1}{4}$

Combine like terms on the left side of the equation.
$\left(-\frac{1}{14} + \frac{1}{16}\right)y + \frac{1}{2} = \frac{1}{4}$

Subtract $\frac{1}{2}$ from both sides of the equation.
$\left(-\frac{1}{14} + \frac{1}{16}\right)y = \frac{1}{4} - \frac{1}{2}$

implify the right side of the equation.
$\left(-\frac{1}{14} + \frac{1}{16}\right)y = -\frac{1}{4}$

Multiply both sides of the equation by $-112$ to solve for $y$ .
$y = -\frac{}{4} \times -112$

Calculate the value of $y$ .
$y =28$

Now that we have the value of $y$ , we can substitute it into the first equation to solve for $x$ .
$x = \frac{28}{7} -$

Calculate the value of $x$ .
$x =4 -$ $x =$ The solution to the system of equations is $x =$ and $y =28$ .