Math / Algebra

Question $\begin{array}{l}6 x+y=8 \\ 3 x-2 y=14\end{array}$

Studdy Solution

STEP 1

What is this asking? We've got two equations with two mystery numbers, x and y, and our mission is to find what those numbers are! Watch out! Be super careful with those negative signs; they can be sneaky!

STEP 2

1. Set up for elimination
2. Eliminate y
3. Solve for x
4. Substitute and solve for y

STEP 3

Let's multiply that first equation, $6x + y = 8$ , by $\bf{2}$ .
Why $\bf{2}$ ?
Because we want to match up the y terms so they cancel out later!
This gives us a shiny new equation: $12x + 2y = 16$ .

STEP 4

Now, let's add our new equation, $12x + 2y = 16$ , to the second equation, $3x - 2y = 14$ .
Watch the magic happen! $\begin{array}{rcr} 12x & +2y & =16 \\ 3x & -2y & =14 \\ 15x & & = 30 \end{array}$ The y terms add to zero!
We're left with $15x = 30$ !

STEP 5

Time to isolate x!
We've got $15x = 30$ .
Let's divide both sides by $\bf{15}$ to get x all by itself. $\frac{15x}{15} = \frac{30}{15}$ $x = \bf{2}$ Woohoo! We found x!

STEP 6

Almost there!
Let's plug our newly discovered x value, $\bf{2}$ , back into the first original equation, $6x + y = 8$ . $6 \cdot (\bf{2}) + y = 8$ $12 + y = 8$

STEP 7

Now, we want y by itself.
Let's subtract $\bf{12}$ from both sides of the equation: $12 + y - 12 = 8 - 12$ $y = \bf{-4}$ Yes! We found y!

STEP 8

So, our dynamic duo is $x = \bf{2}$ and $y = \bf{-4}$ !

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Question $\begin{array}{l}6 x+y=8 \\ 3 x-2 y=14\end{array}$

Studdy Solution

STEP 1

What is this asking? We've got two equations with two mystery numbers, x and y, and our mission is to find what those numbers are! Watch out! Be super careful with those negative signs; they can be sneaky!

STEP 2

1. Set up for elimination
2. Eliminate y
3. Solve for x
4. Substitute and solve for y

STEP 3

Let's multiply that first equation, $6x + y = 8$ , by $\bf{2}$ .
Why $\bf{2}$ ?
Because we want to match up the y terms so they cancel out later!
This gives us a shiny new equation: $12x + 2y = 16$ .

STEP 4

Now, let's add our new equation, $12x + 2y = 16$ , to the second equation, $3x - 2y = 14$ .
Watch the magic happen! $\begin{array}{rcr} 12x & +2y & =16 \\ 3x & -2y & =14 \\ 15x & & = 30 \end{array}$ The y terms add to zero!
We're left with $15x = 30$ !

STEP 5

Time to isolate x!
We've got $15x = 30$ .
Let's divide both sides by $\bf{15}$ to get x all by itself. $\frac{15x}{15} = \frac{30}{15}$ $x = \bf{2}$ Woohoo! We found x!

STEP 6

Almost there!
Let's plug our newly discovered x value, $\bf{2}$ , back into the first original equation, $6x + y = 8$ . $6 \cdot (\bf{2}) + y = 8$ $12 + y = 8$

STEP 7

Now, we want y by itself.
Let's subtract $\bf{12}$ from both sides of the equation: $12 + y - 12 = 8 - 12$ $y = \bf{-4}$ Yes! We found y!

STEP 8

So, our dynamic duo is $x = \bf{2}$ and $y = \bf{-4}$ !

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Question6x+y=83x−2y=14\begin{array}{l}6 x+y=8 \\ 3 x-2 y=14\end{array}6x+y=83x−2y=14​

Studdy Solution

STEP 1

What is this asking? We've got two equations with two mystery numbers, *x* and *y*, and our mission is to find what those numbers are! Watch out! Be super careful with those negative signs; they can be sneaky!

STEP 2

1. Set up for elimination2. Eliminate *y*3. Solve for *x*4. Substitute and solve for *y*

STEP 3

Let's **multiply** that first equation, 6x+y=86x + y = 86x+y=8, by 2\bf{2}2.Why 2\bf{2}2?Because we want to match up the *y* terms so they **cancel** out later!This gives us a shiny new equation: 12x+2y=1612x + 2y = 1612x+2y=16.

STEP 4

STEP 5

Time to **isolate** *x*!We've got 15x=3015x = 3015x=30.Let's **divide** both sides by 15\bf{15}15 to get *x* all by itself. 15x15=3015\frac{15x}{15} = \frac{30}{15}1515x​=1530​ x=2x = \bf{2}x=2Woohoo! We found *x*!

STEP 6

Almost there!Let's **plug** our newly discovered *x* value, 2\bf{2}2, back into the first original equation, 6x+y=86x + y = 86x+y=8. 6⋅(2)+y=86 \cdot (\bf{2}) + y = 86⋅(2)+y=8 12+y=812 + y = 812+y=8

STEP 7

Now, we want *y* by itself.Let's **subtract** 12\bf{12}12 from both sides of the equation: 12+y−12=8−1212 + y - 12 = 8 - 1212+y−12=8−12 y=−4y = \bf{-4}y=−4Yes! We found *y*!

STEP 8

So, our dynamic duo is x=2x = \bf{2}x=2 and y=−4y = \bf{-4}y=−4!

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Question $\begin{array}{l}6 x+y=8 \\ 3 x-2 y=14\end{array}$

What is this asking? We've got two equations with two mystery numbers, x and y, and our mission is to find what those numbers are! Watch out! Be super careful with those negative signs; they can be sneaky!

1. Set up for elimination
2. Eliminate y
3. Solve for x
4. Substitute and solve for y

Let's multiply that first equation, $6x + y = 8$ , by $\bf{2}$ .
Why $\bf{2}$ ?
Because we want to match up the y terms so they cancel out later!
This gives us a shiny new equation: $12x + 2y = 16$ .

Time to isolate x!
We've got $15x = 30$ .
Let's divide both sides by $\bf{15}$ to get x all by itself. $\frac{15x}{15} = \frac{30}{15}$ $x = \bf{2}$ Woohoo! We found x!

Almost there!
Let's plug our newly discovered x value, $\bf{2}$ , back into the first original equation, $6x + y = 8$ . $6 \cdot (\bf{2}) + y = 8$ $12 + y = 8$

Now, we want y by itself.
Let's subtract $\bf{12}$ from both sides of the equation: $12 + y - 12 = 8 - 12$ $y = \bf{-4}$ Yes! We found y!

So, our dynamic duo is $x = \bf{2}$ and $y = \bf{-4}$ !