1 min read Skolematematikk 2019-10-29 Start med 32y−3z4−1232y−3z4−12. Nå bruker vi at vi kan gange med et tall (ulikt 00) over og under brøkstreken: 32y=3⋅22y⋅2=64y32y=3⋅22y⋅2=64y −3z4=−3z⋅y4⋅y=−3zy4y−3z4=−3z⋅y4⋅y=−3zy4y −12=−1⋅2y2⋅2y=−2y4y−12=−1⋅2y2⋅2y=−2y4y Dermed er 32y−3z4−12=64y−3zy4y+−2y4y=6−3zy−2y4y=−2y−3zy+64y32y−3z4−12=64y−3zy4y+−2y4y=6−3zy−2y4y=−2y−3zy+64y