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Sagot :
To solve the system of equations:
[tex]\[ \begin{array}{l} 6x - 3y = 3 \\ -2x + 6y = 14 \end{array} \][/tex]
we need to eliminate either the [tex]\(x\)[/tex]-terms or the [tex]\(y\)[/tex]-terms by adding the equations together. Let's determine the necessary multipliers for each case.
### Eliminating the [tex]\(x\)[/tex]-terms:
To eliminate the [tex]\(x\)[/tex]-terms, we want the coefficients of [tex]\(x\)[/tex] in both equations to have the same magnitude but opposite signs. Looking at the coefficients:
- The coefficient of [tex]\(x\)[/tex] in the first equation is [tex]\(6\)[/tex].
- The coefficient of [tex]\(x\)[/tex] in the second equation is [tex]\(-2\)[/tex].
We need to make the magnitude of the coefficients equal, which can be achieved by finding a common multiple. In this case, the least common multiple of 6 and 2 is 6.
To make the [tex]\(x\)[/tex]-terms cancel, we multiply the second equation by [tex]\(3\)[/tex] (since [tex]\(-2 \times 3 = -6\)[/tex]):
[tex]\[ -2x \times 3 = -6x \][/tex]
This makes the coefficients of [tex]\(x\)[/tex] in both equations [tex]\(6\)[/tex] and [tex]\(-6\)[/tex], respectively, allowing them to cancel out when the equations are added together.
So, the number we would multiply the second equation by in order to eliminate the [tex]\(x\)[/tex]-terms is:
[tex]\[ \boxed{3} \][/tex]
### Eliminating the [tex]\(y\)[/tex]-terms:
To eliminate the [tex]\(y\)[/tex]-terms, we want the coefficients of [tex]\(y\)[/tex] in both equations to have the same magnitude but opposite signs. Looking at the coefficients:
- The coefficient of [tex]\(y\)[/tex] in the first equation is [tex]\(-3\)[/tex].
- The coefficient of [tex]\(y\)[/tex] in the second equation is [tex]\(6\)[/tex].
We need to make the magnitude of the coefficients equal, which can be achieved by finding a common multiple. In this case, the least common multiple of 3 and 6 is 6.
To make the [tex]\(y\)[/tex]-terms cancel, we multiply the first equation by [tex]\(2\)[/tex] (since [tex]\(-3 \times 2 = -6\)[/tex]):
[tex]\[ -3y \times 2 = -6y \][/tex]
This makes the coefficients of [tex]\(y\)[/tex] in both equations [tex]\(-6\)[/tex] and [tex]\(6\)[/tex], respectively, allowing them to cancel out when the equations are added together.
So, the number we would multiply the first equation by in order to eliminate the [tex]\(y\)[/tex]-terms is:
[tex]\[ \boxed{2} \][/tex]
Thus, we have determined that to solve the system by elimination, you would multiply the second equation by [tex]\(3\)[/tex] to eliminate the [tex]\(x\)[/tex]-terms and multiply the first equation by [tex]\(2\)[/tex] to eliminate the [tex]\(y\)[/tex]-terms.
[tex]\[ \begin{array}{l} 6x - 3y = 3 \\ -2x + 6y = 14 \end{array} \][/tex]
we need to eliminate either the [tex]\(x\)[/tex]-terms or the [tex]\(y\)[/tex]-terms by adding the equations together. Let's determine the necessary multipliers for each case.
### Eliminating the [tex]\(x\)[/tex]-terms:
To eliminate the [tex]\(x\)[/tex]-terms, we want the coefficients of [tex]\(x\)[/tex] in both equations to have the same magnitude but opposite signs. Looking at the coefficients:
- The coefficient of [tex]\(x\)[/tex] in the first equation is [tex]\(6\)[/tex].
- The coefficient of [tex]\(x\)[/tex] in the second equation is [tex]\(-2\)[/tex].
We need to make the magnitude of the coefficients equal, which can be achieved by finding a common multiple. In this case, the least common multiple of 6 and 2 is 6.
To make the [tex]\(x\)[/tex]-terms cancel, we multiply the second equation by [tex]\(3\)[/tex] (since [tex]\(-2 \times 3 = -6\)[/tex]):
[tex]\[ -2x \times 3 = -6x \][/tex]
This makes the coefficients of [tex]\(x\)[/tex] in both equations [tex]\(6\)[/tex] and [tex]\(-6\)[/tex], respectively, allowing them to cancel out when the equations are added together.
So, the number we would multiply the second equation by in order to eliminate the [tex]\(x\)[/tex]-terms is:
[tex]\[ \boxed{3} \][/tex]
### Eliminating the [tex]\(y\)[/tex]-terms:
To eliminate the [tex]\(y\)[/tex]-terms, we want the coefficients of [tex]\(y\)[/tex] in both equations to have the same magnitude but opposite signs. Looking at the coefficients:
- The coefficient of [tex]\(y\)[/tex] in the first equation is [tex]\(-3\)[/tex].
- The coefficient of [tex]\(y\)[/tex] in the second equation is [tex]\(6\)[/tex].
We need to make the magnitude of the coefficients equal, which can be achieved by finding a common multiple. In this case, the least common multiple of 3 and 6 is 6.
To make the [tex]\(y\)[/tex]-terms cancel, we multiply the first equation by [tex]\(2\)[/tex] (since [tex]\(-3 \times 2 = -6\)[/tex]):
[tex]\[ -3y \times 2 = -6y \][/tex]
This makes the coefficients of [tex]\(y\)[/tex] in both equations [tex]\(-6\)[/tex] and [tex]\(6\)[/tex], respectively, allowing them to cancel out when the equations are added together.
So, the number we would multiply the first equation by in order to eliminate the [tex]\(y\)[/tex]-terms is:
[tex]\[ \boxed{2} \][/tex]
Thus, we have determined that to solve the system by elimination, you would multiply the second equation by [tex]\(3\)[/tex] to eliminate the [tex]\(x\)[/tex]-terms and multiply the first equation by [tex]\(2\)[/tex] to eliminate the [tex]\(y\)[/tex]-terms.
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