To solve this system of linear equations by eliminating the [tex]$y$[/tex]-terms, we need to manipulate the equations so that the coefficients of [tex]$y$[/tex] in both equations are equal in magnitude but opposite in sign. This allows us to add the equations together, thereby eliminating the [tex]$y$[/tex]-terms and solving for [tex]$x$[/tex].
Given the equations:
1. [tex]\(5x - 4y = 28\)[/tex]
2. [tex]\(3x - 9y = 30\)[/tex]
1. First equation: [tex]\(5x - 4y = 28\)[/tex]
- To make the coefficient of [tex]\(y\)[/tex] suitable for elimination, we can multiply this equation by 3. This results in:
[tex]\[3 \cdot (5x - 4y) = 3 \cdot 28\][/tex]
[tex]\[15x - 12y = 84\][/tex]
2. Second equation: [tex]\(3x - 9y = 30\)[/tex]
- To make the coefficients of [tex]\(y\)[/tex] in the two equations opposite in sign, we should multiply this equation by -5. This results in:
[tex]\[-5 \cdot (3x - 9y) = -5 \cdot 30\][/tex]
[tex]\[-15x + 45y = -150\][/tex]
Now, we have two new equations:
1. [tex]\(15x - 12y = 84\)[/tex]
2. [tex]\(-15x + 45y = -150\)[/tex]
By adding these two equations, we effectively eliminate the [tex]$y$[/tex]-terms:
[tex]\[
(15x - 12y) + (-15x + 45y) = 84 + (-150)
\][/tex]
[tex]\[(15x - 15x) + (-12y + 45y) = 84 - 150\][/tex]
[tex]\[0x + 33y = -66\][/tex]
However, the question was specifically about the constants by which the original equations should be multiplied before adding them together to eliminate the [tex]$y$[/tex]-terms. Therefore, the constants are:
- The first equation should be multiplied by [tex]\(3\)[/tex]
- The second equation should be multiplied by [tex]\(-5\)[/tex]
Thus, the correct answer is:
- The first equation should be multiplied by 3 and the second equation by -5.
