Question

pear assessment
6d5 exit ticket
created by kristine ingbernan
1
a system of equations is shown.
6x - 3y = 15
-3x + 9y = -15
in order to solve this system by elimination, first a by b. then, when you add the equations together, the c term will cancel.
a

• multiplies the first equation
• multiplies the second equation

b
3
2
1
-1
c
x
y

Explanation:

Step1: Analyze the equations

We have the system:
\(6x - 3y = 15\)
\(-3x - 9y = -18\)

To solve by elimination, we want to make the coefficients of one variable opposites or equal. Let's focus on the \(x\) or \(y\) terms. Let's look at the \(x\) coefficients: \(6\) and \(-3\). If we multiply the second equation by \(2\), the \(x\)-coefficient becomes \(-6\), which is the opposite of \(6\) in the first equation.

Step2: Multiply the second equation

Multiply the second equation \(-3x - 9y = -18\) by \(2\):
\(2\times(-3x - 9y)=2\times(-18)\)
\(-6x - 18y = -36\)

Now our system is:
\(6x - 3y = 15\)
\(-6x - 18y = -36\)

If we add these two equations, the \(x\) terms will cancel:
\((6x - 3y)+(-6x - 18y)=15+(-36)\)
\(6x - 3y - 6x - 18y = 15 - 36\)
\(-21y = -21\)
\(y = 1\)

Then substitute \(y = 1\) back into the first equation \(6x - 3(1)=15\)
\(6x - 3 = 15\)
\(6x = 18\)
\(x = 3\)

But the question is about what to multiply to eliminate. Let's check the \(y\) terms: first equation has \(-3y\), second has \(-9y\). If we multiply the first equation by \(3\), we get \(18x - 9y = 45\), and the second equation is \(-3x - 9y = -18\). Then subtracting would eliminate \(y\). But the options: let's see the first blank (A: multiplies the first equation, B: multiplies the second equation), second blank (the number), third blank (the term that cancels).

Wait, the original equations: \(6x - 3y = 15\) and \(-3x - 9y = -18\). Let's look at the \(x\) coefficients: \(6\) and \(-3\). If we multiply the second equation by \(2\), we get \(-6x - 18y = -36\), then adding the first equation (\(6x - 3y = 15\)) will eliminate \(x\) (since \(6x + (-6x)=0\)). So:

First blank: B? Wait no, A is multiplies the first equation, B is multiplies the second equation. Wait, to eliminate \(x\), we need \(6x\) and \(-6x\). So multiply the second equation (which has \(-3x\)) by \(2\) to get \(-6x\). So the first blank: B (multiplies the second equation), second blank: \(2\), third blank: \(x\) (since \(6x + (-6x)=0\)).

Let's check the options:

A: multiplies the first equation

B: multiplies the second equation

Numbers: 2, -2, 3, -3

Terms: x, y

So:

To solve by elimination, we can [B: multiplies the second equation] by [2], then, when we add the equations, the [x] terms will cancel.

Let's verify:

Second equation: \(-3x - 9y = -18\) multiplied by 2: \(-6x - 18y = -36\)

First equation: \(6x - 3y = 15\)

Add them: \(6x - 3y -6x -18y = 15 -36\) → \(-21y = -21\) → \(y=1\). Wait, no, the \(x\) terms cancel (6x -6x=0). So the term that cancels is \(x\).

So the answers:

First blank: B (multiplies the second equation)

Second blank: 2

Third blank: x

But let's check the original problem's options again (from the image):

A: multiplies the first equation

B: multiplies the second equation

Numbers: 2, -2, 3, -3

Terms: x, y

So putting it together:

In order to solve this system by elimination, you [B: multiplies the second equation] by [2]. Then, when you add the equations together, the [x] terms will cancel.

Let's check another way: if we multiply the first equation by 3: \(18x -9y = 45\), second equation: \(-3x -9y = -18\). Subtract the second from the first: \(21x = 63\) → \(x=3\), then \(y=1\). But that eliminates \(y\) (since -9y - (-9y)=0? No, -9y - (-9y)=0? Wait, first equation after multiplying by 3: 18x -9y=45, second: -3x -9y=-18. Subtract: (18x -9y) - (-3x -9y)=45 - (-18) → 21x=63 → x=3. So that eliminates \(y\) (because -9y - (-9y)=0). But the options for the term that cancels: x or y.

Wait, the original problem's equations: \(6x - 3y = 15\) and \(-3x - 9y = -18\). Let's list the coefficients:

For \(x\): 6 and -3

For \(y\): -3 and -9

To eliminate \(x\): make 6x and -6x. So multiply second equation by 2: -6x -18y = -36. Then add to first equation: 6x -3y -6x -18y = 15 -36 → -21y = -21 → y=1. Here, x terms cancel.

To eliminate \(y\): make -3y and 9y (or -9y and 3y). Multiply first equation by 3: 18x -9y = 45. Second equation: -3x -9y = -18. Subtract: (18x -9y) - (-3x -9y)=45 - (-18) → 21x=63 → x=3. Here, y terms cancel (-9y - (-9y)=0? No, -9y - (-9y)=0? Wait, 18x -9y - (-3x -9y)=18x -9y +3x +9y=21x. So the y terms (-9y +9y) cancel. So in this case, y terms cancel.

But the options: let's see the numbers. If we multiply the first equation by 3, number is 3, term is y. If we multiply the second equation by 2, number is 2, term is x.

Looking at the options for the number: 2, -2, 3, -3.

Let's check the problem again. The user's image:

"A system of equations is shown.

\(6x - 3y = 15\)

\(-3x - 9y = -18\)

In order to solve this system by elimination, you [blank1] by [blank2]. Then, when you add the equations together, the [blank3] terms will cancel.

A: multiplies the first equation

B: multiplies the second equation

Numbers: 2, -2, 3, -3

Terms: x, y"

So let's find which combination works.

Case 1: Eliminate x.

Multiply second equation (B) by 2:

Second equation: \(-3x -9y = -18\) *2 → \(-6x -18y = -36\)

First equation: \(6x -3y = 15\)

Add: \(6x -3y -6x -18y = 15 -36\) → -21y = -21 → y=1. Here, x terms cancel. So blank1: B, blank2: 2, blank3: x.

Case 2: Eliminate y.

Multiply first equation (A) by 3:

First equation: \(6x -3y =15\) *3 → \(18x -9y =45\)

Second equation: \(-3x -9y = -18\)

Subtract: \(18x -9y - (-3x -9y)=45 - (-18)\) → \(21x =63\) → x=3. Here, y terms cancel. So blank1: A, blank2: 3, blank3: y.

Now, check the number options: 2, -2, 3, -3. So 2 and 3 are options.

Now, let's see the original equations. The y coefficients are -3 and -9. The LCM of 3 and 9 is 9. So to make them equal (for elimination by addition or subtraction), if we want to eliminate y, we can multiply the first equation by 3 (to get -9y) and the second equation has -9y, then subtract to eliminate y.

Or multiply the second equation by 1/3, but that's not an option. So the options have 2,3,-2,-3.

Let's check the x coefficients: 6 and -3. LCM of 6 and 3 is 6. So multiply the second equation by 2 (to get -6x) and the first equation has 6x, then add to eliminate x.

So the correct choice is:

blank1: B (multiplies the second equation)

blank2: 2

blank3: x

So the answers:

First blank: B

Second blank: 2

Third blank: x

But let's confirm with the options:

A: multiplies the first equation

B: multiplies the second equation

Numbers: 2, -2, 3, -3

Terms: x, y

So putting it together:

In order to solve this system by elimination, you [B: multiplies the second equation] by [2]. Then, when you add the equations together, the [x] terms will cancel.

Let's verify the addition:

First equation: \(6x - 3y = 15\)

Second equation multiplied by 2: \(-6x - 18y = -36\)

Add them: \(6x - 3y -6x -18y = 15 -36\)

Simplify: \((6x -6x) + (-3y -18y) = -21\)

\(0x -21y = -21\)

So the x terms (6x and -6x) cancel. Correct.

So the answers are:

First blank: B

Second blank: 2

Third blank: x

Answer:

First blank: B. multiplies the second equation
Second blank: 2
Third blank: x