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Sagot :
To find the solution to the system of equations:
[tex]\[ \begin{array}{l} y=-\frac{1}{2} x+9 \\ y=x+7 \end{array} \][/tex]
we need to determine the point where these two lines intersect.
1. Set the equations equal to each other because [tex]\( y \)[/tex] is the same in both equations at the point of intersection:
[tex]\[ -\frac{1}{2} x + 9 = x + 7 \][/tex]
2. Solve for [tex]\( x \)[/tex]:
- Move [tex]\( x \)[/tex] terms to one side of the equation:
[tex]\[ 9 - 7 = x + \frac{1}{2} x \][/tex]
- Simplify the equation:
[tex]\[ 2 = \frac{3}{2} x \][/tex]
- Multiply both sides by [tex]\(\frac{2}{3}\)[/tex] to isolate [tex]\( x \)[/tex]:
[tex]\[ x = \frac{2 \times 2}{3} = \frac{4}{3} \][/tex]
3. Solve for [tex]\( y \)[/tex] using one of the original equations, substituting [tex]\( x = \frac{4}{3} \)[/tex]:
[tex]\[ y = x + 7 \\ y = \frac{4}{3} + 7 \][/tex]
- Convert 7 to a fraction with a denominator of 3:
[tex]\[ y = \frac{4}{3} + \frac{21}{3} = \frac{25}{3} \][/tex]
Thus, the point of intersection is:
[tex]\[ \left( \frac{4}{3}, \frac{25}{3} \right) \][/tex]
Expressing this in decimal form for clarity:
[tex]\[ \left( 1.33333333333333, 8.33333333333333 \right) \][/tex]
Given the options:
- Lines [tex]\(y=-\frac{1}{2} x+9\)[/tex] and [tex]\(y=x+7\)[/tex] intersect the [tex]\(x\)[/tex]-axis.
- Lines [tex]\(y=-\frac{1}{2} x+9\)[/tex] and [tex]\(y=x+7\)[/tex] intersect the [tex]\(y\)[/tex]-axis.
- Line [tex]\(y=-\frac{1}{2} x+9\)[/tex] intersects the origin.
- Line [tex]\(y=-\frac{1}{2} x+9\)[/tex] intersects line [tex]\(y=x+7\)[/tex].
The best description of this solution is:
Line [tex]\( y = -\frac{1}{2} x + 9 \)[/tex] intersects line [tex]\( y = x + 7 \)[/tex].
[tex]\[ \begin{array}{l} y=-\frac{1}{2} x+9 \\ y=x+7 \end{array} \][/tex]
we need to determine the point where these two lines intersect.
1. Set the equations equal to each other because [tex]\( y \)[/tex] is the same in both equations at the point of intersection:
[tex]\[ -\frac{1}{2} x + 9 = x + 7 \][/tex]
2. Solve for [tex]\( x \)[/tex]:
- Move [tex]\( x \)[/tex] terms to one side of the equation:
[tex]\[ 9 - 7 = x + \frac{1}{2} x \][/tex]
- Simplify the equation:
[tex]\[ 2 = \frac{3}{2} x \][/tex]
- Multiply both sides by [tex]\(\frac{2}{3}\)[/tex] to isolate [tex]\( x \)[/tex]:
[tex]\[ x = \frac{2 \times 2}{3} = \frac{4}{3} \][/tex]
3. Solve for [tex]\( y \)[/tex] using one of the original equations, substituting [tex]\( x = \frac{4}{3} \)[/tex]:
[tex]\[ y = x + 7 \\ y = \frac{4}{3} + 7 \][/tex]
- Convert 7 to a fraction with a denominator of 3:
[tex]\[ y = \frac{4}{3} + \frac{21}{3} = \frac{25}{3} \][/tex]
Thus, the point of intersection is:
[tex]\[ \left( \frac{4}{3}, \frac{25}{3} \right) \][/tex]
Expressing this in decimal form for clarity:
[tex]\[ \left( 1.33333333333333, 8.33333333333333 \right) \][/tex]
Given the options:
- Lines [tex]\(y=-\frac{1}{2} x+9\)[/tex] and [tex]\(y=x+7\)[/tex] intersect the [tex]\(x\)[/tex]-axis.
- Lines [tex]\(y=-\frac{1}{2} x+9\)[/tex] and [tex]\(y=x+7\)[/tex] intersect the [tex]\(y\)[/tex]-axis.
- Line [tex]\(y=-\frac{1}{2} x+9\)[/tex] intersects the origin.
- Line [tex]\(y=-\frac{1}{2} x+9\)[/tex] intersects line [tex]\(y=x+7\)[/tex].
The best description of this solution is:
Line [tex]\( y = -\frac{1}{2} x + 9 \)[/tex] intersects line [tex]\( y = x + 7 \)[/tex].
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