Q. Find the x-intercept of the line \(5x + 11y = -2\).

Q: What is the x-intercept of the line 5x+11y=-2?.

Set y=0. Then 5x=-2, so x=-\tfrac{2}{5}. The x-intercept is (-\tfrac{2}{5},0)..

Q: How do I find the y-intercept of 5x+11y=-2?

Set x=0. Then 11y=-2, so y=-\tfrac{2}{11}. The y-intercept is (0,-\tfrac{2}{11}).

Q: How do I get the slope from 5x+11y=-2?.

Solve for y: 11y=-5x-2 so y=-\tfrac{5}{11}x-\tfrac{2}{11}. The slope is -\tfrac{5}{11}.

Q: How can I graph this line using intercepts?

Plot the x-intercept \left(- \tfrac{2}{5},0\right) and the y-intercept \left(0,-\tfrac{2}{11}\right), then draw the straight line through those two points..

Q: How do I check my x-intercept is correct?

Substitute (- \tfrac{2}{5},0) into 5x+11y : 5(-\tfrac{2}{5})+11(0)=-2 . The equation holds, so the intercept is correct.

Answer

Set \(y=0\). Then \(5x+11(0)=-2\), so \(5x=-2\) and \(x=-\tfrac{2}{5}\). The x-intercept is \(\left(-\tfrac{2}{5},\,0\right)\).

Detailed Explanation

Find the x-intercept of the line

Given the equation \(5x + 11y = -2\).

Recall the definition of an x-intercept.The x-intercept is the point where the graph crosses the x-axis. At any point on the x-axis the y-coordinate is 0. Therefore to find the x-intercept we set \(y = 0\) in the equation and solve for \(x\).
Substitute \(y = 0\) into the equation.Replace \(y\) with \(0\) in \(5x + 11y = -2\). This gives
\(5x + 11\cdot 0 = -2\).
Simplify the equation after substitution.Since \(11\cdot 0 = 0\), the equation becomes
\(5x = -2\).
Solve for \(x\).Divide both sides of \(5x = -2\) by \(5\) to isolate \(x\):
\(x = -\tfrac{2}{5}\).

As a decimal, \(x = -0.4\).
Write the x-intercept as a point.The x-intercept is the point with coordinates \(\left(-\tfrac{2}{5},\,0\right)\).

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Algebra FAQs

What is the x-intercept of the line \(5x+11y=-2\)?.

Set \(y=0\). Then \(5x=-2\), so \(x=-\tfrac{2}{5}\). The x-intercept is \((-\tfrac{2}{5},0)\)..

Why do we set \(y=0\) to find the x-intercept?.

The x-intercept is the point where the graph crosses the x-axis, and every point on the x-axis has \(y=0\). Substitute \(y=0\) and solve for \(x\).

How do I find the y-intercept of \(5x+11y=-2\)?

Set \(x=0\). Then \(11y=-2\), so \(y=-\tfrac{2}{11}\). The y-intercept is \((0,-\tfrac{2}{11})\).

How do I get the slope from \(5x+11y=-2\)?.

Solve for \(y\): \(11y=-5x-2\) so \(y=-\tfrac{5}{11}x-\tfrac{2}{11}\). The slope is \(-\tfrac{5}{11}\).

How can I graph this line using intercepts?

Plot the x-intercept \(\left(- \tfrac{2}{5},0\right)\) and the y-intercept \(\left(0,-\tfrac{2}{11}\right)\), then draw the straight line through those two points..

What is the intercept form and how does this line fit it?

How do I check my x-intercept is correct?

Substitute \( (- \tfrac{2}{5},0) \) into \( 5x+11y \): \( 5(-\tfrac{2}{5})+11(0)=-2 \). The equation holds, so the intercept is correct.

What does a negative \(x\)-intercept mean geometrically?

Negative x-intercept means the line crosses the x-axis to the left of the origin; here it crosses at \(x=-\tfrac{2}{5}\), so left of \(0\).

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