Q. Find the y-intercept of the line \(15x + 18y = -20\).
Answer
Set x=0: \(18y=-20\), so \(y=-20/18=-10/9\). Thus the y-intercept is \((0,-10/9)\).
Detailed Explanation
Step 1 — What the y-intercept means
The y-intercept is the point where the line crosses the y-axis, so the x-coordinate is 0. Therefore set \(x = 0\).
Step 2 — Substitute x = 0 into the equation
Start from the given equation:
\[15x + 18y = -20\]
Substitute \(x = 0\):
\[15(0) + 18y = -20\]
Compute \(15(0) = 0\) to get:
\[0 + 18y = -20\]
So
\[18y = -20\]
Step 3 — Solve for y
Divide both sides by 18 to isolate \(y\):
\[y = \frac{-20}{18}\]
Reduce the fraction by dividing numerator and denominator by 2:
\[y = \frac{-10}{9}\]
Step 4 — State the y-intercept
The y-intercept is the point on the y-axis corresponding to this y-value:
\[(0, -\tfrac{10}{9})\]
See full solution
Algebra FAQs
How do I find the y-intercept of the line \(15x+18y=-20\)?.
Set \(x=0\). Then \(18y=-20\), so \(y=-\tfrac{10}{9}\). The y-intercept point is \((0,-\tfrac{10}{9})\).
What is the slope of this line?
Solve for \(y\): \(y=-\tfrac{5}{6}x-\tfrac{10}{9}\). The slope is \(-\tfrac{5}{6}\).
What is the \(x\)-intercept of the line?
Set \(y=0\). Then \(15x=-20\), so \(x=-\tfrac{4}{3}\). The x-intercept is \((-\tfrac{4}{3},0)\).
How can I graph the line quickly?
Plot the intercepts \( (0,-\tfrac{10}{9}) \) and \( (-\tfrac{4}{3},0) \), then draw the line through them; or use slope \(-\tfrac{5}{6}\) from the y-intercept: down 5, right 6.
How do I write the equation in slope-intercept form?
Rearrange: \(y=-\tfrac{5}{6}x-\tfrac{10}{9}\).
What is the \(y\)-intercept as a decimal?
What is the \(y\)-intercept as a decimal?
Can the intercepts be simplified or scaled for nicer integers?
Multiply by a common factor: for example, scaling both coordinates by 9 gives y-intercept \( (0,-10) \) in scaled units, but the true intercept remains \( (0,-\tfrac{10}{9}) \).
Why is setting \(x=0\) the correct method for y-intercept?
The y-intercept is where the line crosses the y-axis, and every point on the y-axis has \(x=0\); solving with \(x=0\) gives the y-coordinate of that crossing.
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