Direct, Inverse, and Joint Variation Practice

Translate as y=kx, solve for k from the matched pair, then use the new x-value.

From \(y=kx\), the known pair gives \(k=2\). Therefore \(y=2(7)=14\).

Translate as y=kx, solve for k from the matched pair, then use the new x-value.

From \(y=kx\), the known pair gives \(k=3\). Therefore \(y=3(9)=27\).

Translate as y=kx, solve for k from the matched pair, then use the new x-value.

From \(y=kx\), the known pair gives \(k=4\). Therefore \(y=4(8)=32\).

Translate as y=kx, solve for k from the matched pair, then use the new x-value.

From \(y=kx\), the known pair gives \(k=5\). Therefore \(y=5(11)=55\).

Translate as y=kx, solve for k from the matched pair, then use the new x-value.

From \(y=kx\), the known pair gives \(k=6\). Therefore \(y=6(7)=42\).

Translate as y=kx, solve for k from the matched pair, then use the new x-value.

From \(y=kx\), the known pair gives \(k=7\). Therefore \(y=7(6)=42\).

Translate as y=kx, solve for k from the matched pair, then use the new x-value.

From \(y=kx\), the known pair gives \(k=8\). Therefore \(y=8(5)=40\).

Translate as y=kx, solve for k from the matched pair, then use the new x-value.

From \(y=kx\), the known pair gives \(k=9\). Therefore \(y=9(8)=72\).

Translate as y=kx, solve for k from the matched pair, then use the new x-value.

From \(y=kx\), the known pair gives \(k=3\). Therefore \(y=3(12)=36\).

Translate as y=kx, solve for k from the matched pair, then use the new x-value.

From \(y=kx\), the known pair gives \(k=4\). Therefore \(y=4(13)=52\).

Translate as y=kx, solve for k from the matched pair, then use the new x-value.

From \(y=kx\), the known pair gives \(k=5\). Therefore \(y=5(9)=45\).

Translate as y=kx, solve for k from the matched pair, then use the new x-value.

From \(y=kx\), the known pair gives \(k=6\). Therefore \(y=6(11)=66\).

Translate as y=kx, solve for k from the matched pair, then use the new x-value.

From \(y=kx\), the known pair gives \(k=7\). Therefore \(y=7(10)=70\).

Translate as y=kx, solve for k from the matched pair, then use the new x-value.

From \(y=kx\), the known pair gives \(k=8\). Therefore \(y=8(9)=72\).

Translate as y=kx, solve for k from the matched pair, then use the new x-value.

From \(y=kx\), the known pair gives \(k=9\). Therefore \(y=9(7)=63\).

Translate as y=kx, solve for k from the matched pair, then use the new x-value.

From \(y=kx\), the known pair gives \(k=10\). Therefore \(y=10(8)=80\).

Translate as y=kx, solve for k from the matched pair, then use the new x-value.

From \(y=kx\), the known pair gives \(k=-2\). Therefore \(y=-2(9)=-18\).

Translate as y=kx, solve for k from the matched pair, then use the new x-value.

From \(y=kx\), the known pair gives \(k=-3\). Therefore \(y=-3(8)=-24\).

Translate as y=kx, solve for k from the matched pair, then use the new x-value.

From \(y=kx\), the known pair gives \(k=11\). Therefore \(y=11(6)=66\).

Translate as y=kx, solve for k from the matched pair, then use the new x-value.

From \(y=kx\), the known pair gives \(k=12\). Therefore \(y=12(7)=84\).

Use the constant product xy=k; do not use the direct-variation quotient.

Inverse variation keeps \(xy=k\). The known pair gives \(k=24\), so \(y=24/6=4\).

Use the constant product xy=k; do not use the direct-variation quotient.

Inverse variation keeps \(xy=k\). The known pair gives \(k=30\), so \(y=30/10=3\).

Use the constant product xy=k; do not use the direct-variation quotient.

Inverse variation keeps \(xy=k\). The known pair gives \(k=36\), so \(y=36/9=4\).

Use the constant product xy=k; do not use the direct-variation quotient.

Inverse variation keeps \(xy=k\). The known pair gives \(k=40\), so \(y=40/8=5\).

Use the constant product xy=k; do not use the direct-variation quotient.

Inverse variation keeps \(xy=k\). The known pair gives \(k=42\), so \(y=42/7=6\).

Use the constant product xy=k; do not use the direct-variation quotient.

Inverse variation keeps \(xy=k\). The known pair gives \(k=48\), so \(y=48/12=4\).

Use the constant product xy=k; do not use the direct-variation quotient.

Inverse variation keeps \(xy=k\). The known pair gives \(k=54\), so \(y=54/9=6\).

Use the constant product xy=k; do not use the direct-variation quotient.

Inverse variation keeps \(xy=k\). The known pair gives \(k=60\), so \(y=60/12=5\).

Use the constant product xy=k; do not use the direct-variation quotient.

Inverse variation keeps \(xy=k\). The known pair gives \(k=72\), so \(y=72/9=8\).

Use the constant product xy=k; do not use the direct-variation quotient.

Inverse variation keeps \(xy=k\). The known pair gives \(k=80\), so \(y=80/16=5\).

Use the constant product xy=k; do not use the direct-variation quotient.

Inverse variation keeps \(xy=k\). The known pair gives \(k=90\), so \(y=90/15=6\).

Use the constant product xy=k; do not use the direct-variation quotient.

Inverse variation keeps \(xy=k\). The known pair gives \(k=96\), so \(y=96/16=6\).

Use the constant product xy=k; do not use the direct-variation quotient.

Inverse variation keeps \(xy=k\). The known pair gives \(k=108\), so \(y=108/18=6\).

Use the constant product xy=k; do not use the direct-variation quotient.

Inverse variation keeps \(xy=k\). The known pair gives \(k=120\), so \(y=120/20=6\).

Use the constant product xy=k; do not use the direct-variation quotient.

Inverse variation keeps \(xy=k\). The known pair gives \(k=144\), so \(y=144/24=6\).

Include every jointly named variable, solve for k, and then substitute the complete new pair.

The model is \(z=kxy\), and the known triple gives \(k=1\). Thus \(z=1(4)(5)=20\).

Include every jointly named variable, solve for k, and then substitute the complete new pair.

The model is \(z=kxy\), and the known triple gives \(k=2\). Thus \(z=2(3)(5)=30\).

Include every jointly named variable, solve for k, and then substitute the complete new pair.

The model is \(z=kxy\), and the known triple gives \(k=3\). Thus \(z=3(2)(6)=36\).

Include every jointly named variable, solve for k, and then substitute the complete new pair.

The model is \(z=kxy\), and the known triple gives \(k=2\). Thus \(z=2(5)(2)=20\).

Include every jointly named variable, solve for k, and then substitute the complete new pair.

The model is \(z=kxy\), and the known triple gives \(k=4\). Thus \(z=4(5)(4)=80\).

Include every jointly named variable, solve for k, and then substitute the complete new pair.

The model is \(z=kxy\), and the known triple gives \(k=5\). Thus \(z=5(3)(4)=60\).

Include every jointly named variable, solve for k, and then substitute the complete new pair.

The model is \(z=kxy\), and the known triple gives \(k=3\). Thus \(z=3(4)(3)=36\).

Include every jointly named variable, solve for k, and then substitute the complete new pair.

The model is \(z=kxy\), and the known triple gives \(k=2\). Thus \(z=2(6)(2)=24\).

Include every jointly named variable, solve for k, and then substitute the complete new pair.

The model is \(z=kxy\), and the known triple gives \(k=6\). Thus \(z=6(2)(5)=60\).

Include every jointly named variable, solve for k, and then substitute the complete new pair.

The model is \(z=kxy\), and the known triple gives \(k=4\). Thus \(z=4(5)(3)=60\).

Look for a constant multiple of x with no added intercept.

The equation has form \(y=kx\) and passes through the origin.

Test xy for inverse variation and y/x for direct variation.

Each row has constant product \(xy=24\), so \(y=24/x\).

Translate the power before applying the scale factor.

The model is \(y=kx^2\); replacing x by 3x multiplies y by \(3^2=9\).

Apply inverse behavior to the entire stated power.

The model \(y=k/x^2\) makes doubling x multiply y by \(1/2^2=1/4\).

Check the intercept as well as linearity.

Direct variation must have form \(y=kx\), so its vertical intercept must be zero.